• 2017"百度之星"程序设计大赛

    度度熊保护村庄

    Accepts: 13
    Submissions: 488
    Time Limit: 2000/1000 MS (Java/Others)
    Memory Limit: 32768/32768 K (Java/Others)
    Problem Description

    哗啦啦村袭击了喵哈哈村！

    度度熊为了拯救喵哈哈村，带着自己的伙伴去救援喵哈哈村去了！度度熊与伙伴们很快的就过来占据了喵哈哈村的各个军事要地，牢牢的守住了喵哈哈村。

    但是度度熊发现，这是一场旷日持久的战斗，所以度度熊决定要以逸待劳，保存尽量多的体力，去迎战哗啦啦村的战士。

    于是度度熊决定派尽量多的人去休息，但是同时也不能松懈对喵哈哈村的保护。

    换句话而言，度度熊希望尽量多的人休息，而且存在一个包围圈由剩下的人组成，且能够恰好的包围住喵哈哈村的所有住房（包括边界）。

    请问最多能让多少个人休息呢？

    Input

    本题包含若干组测试数据。

    第一行一个整数n，表示喵哈哈村的住房数量。

    接下来n行，每行两个整数(x1[i],y1[i])，表示喵哈哈村的住房坐标。

    第n+1行一个整数m，表示度度熊的士兵数量。

    接下来m行，每行两个整数(x2[i],y2[i])，表示度度熊伙伴的坐标。

    满足:

    1<=n,m<=500

    -10000<=x1[i],x2[i],y1[i],y2[i]<=10000

    Output

    请输出最多的人员休息的数目。

    如果无法保护整个村庄的话，输出"ToT"

    Sample Input
    2
1 1
2 2
4
0 0
0 4
4 2
4 0
1
1 1
2
0 0
0 1
    Sample Output
    1
ToT

    题目链接：http://bestcoder.hdu.edu.cn/contests/contest_showproblem.php?cid=774&pid=1001

    分析：参考 BZOJ1027，floyd求最小环，有两个情况要特判，一个是重点，一个室房屋恰好在两点的连线上

    O(n*n)暴力枚举所有的点对，对于每个点O(n)检测，如果所有的房子都在一条连线的一侧，则这两个点连线，否则不连，如果这个图中都不存在环，那么输出ToT
    下面给出AC代码：
     1 #include<stdio.h>
 2 #include<string.h>
 3 #include<algorithm>
 4 using namespace std;
 5 #define inf 1044266558
 6 typedef struct Point
 7 {
 8     int x, y;
 9     Point operator - ( const Point &b ) const
10     {
11         Point c;
12         c.x = x-b.x; c.y = y-b.y;
13         return c;
14     }
15     double operator * ( const Point &b ) const
16     {
17         return x*b.y-y*b.x;
18     }
19 }Point;
20 Point h[505], s[505];
21 int n, m, ans, road[505][505];
22 bool Jud(Point x, Point y, Point z)
23 {
24     if((x.x<z.x && y.x<z.x) || (x.y<z.y && y.y<z.y) || (x.x>z.x && y.x>z.x) || (x.y>z.y && y.y>z.y))
25         return 1;
26     return 0;
27 }
28 int main(void)
29 {
30     int i, j, k, flag;
31     while(scanf("%d", &n)!=EOF)
32     {
33         memset(road, 62, sizeof(road));
34         for(i=1;i<=n;i++)
35             scanf("%d%d", &h[i].x, &h[i].y);
36         scanf("%d", &m);
37         for(i=1;i<=m;i++)
38             scanf("%d%d", &s[i].x, &s[i].y);
39         for(i=1;i<=m;i++)
40         {
41             for(j=1;j<=m;j++)
42             {
43                 flag = 1;
44                 for(k=1;k<=n;k++)
45                 {
46                     if((s[i]-s[j])*(s[i]-h[k])<0 || (s[i]-s[j])*(s[i]-h[k])==0 && Jud(s[i], s[j], h[k]))
47                     {
48                         flag = 0;
49                         break;
50                     }
51                 }
52                 if(flag)
53                     road[i][j] = 1;
54             }
55         }
56         ans = inf;
57         for(k=1;k<=m;k++)
58         {
59             for(i=1;i<=m;i++)
60             {
61                 if(road[i][k]==inf)
62                     continue;
63                 for(j=1;j<=m;j++)
64                     road[i][j] = min(road[i][j], road[i][k]+road[k][j]);
65             }
66         }
67         for(i=1;i<=m;i++)
68             ans = min(ans, road[i][i]);
69         if(ans>m)
70             printf("ToT
");
71         else
72             printf("%d
", m-ans);
73     }
74     return 0;    
75 }

    度度熊的王国战略

    Accepts: 173
    Submissions: 3298
    Time Limit: 40000/20000 MS (Java/Others)
    Memory Limit: 32768/132768 K (Java/Others)
    Problem Description

    度度熊国王率领着喵哈哈族的勇士，准备进攻哗啦啦族。

    哗啦啦族是一个强悍的民族，里面有充满智慧的谋士，拥有无穷力量的战士。

    所以这一场战争，将会十分艰难。

    为了更好的进攻哗啦啦族，度度熊决定首先应该从内部瓦解哗啦啦族。

    第一步就是应该使得哗啦啦族内部不能同心齐力，需要内部有间隙。

    哗啦啦族一共有n个将领，他们一共有m个强关系，摧毁每一个强关系都需要一定的代价。

    现在度度熊命令你需要摧毁一些强关系，使得内部的将领，不能通过这些强关系，连成一个完整的连通块，以保证战争的顺利进行。

    请问最少应该付出多少的代价。

    Input

    本题包含若干组测试数据。

    第一行两个整数n，m，表示有n个将领，m个关系。

    接下来m行，每行三个整数u,v,w。表示u将领和v将领之间存在一个强关系，摧毁这个强关系需要代价w

    数据范围：

    2<=n<=3000

    1<=m<=100000

    1<=u,v<=n

    1<=w<=1000

    Output

    对于每组测试数据，输出最小需要的代价。

    Sample Input
    2 1
1 2 1
3 3
1 2 5
1 2 4
2 3 3
    Sample Output
    1
3

    题目链接：http://bestcoder.hdu.edu.cn/contests/contest_showproblem.php?cid=774&pid=1002

    分析：歪解(并查集可过；正解似乎是堆优化+SW(最小正割)，啥玩意，不懂！

    下面给出AC代码：

      1 #include <bits/stdc++.h>
  2 using namespace std;
  3 typedef long long ll;
  4 #define max(a,b) (a)>(b)?(a):(b);
  5 #define min(a,b) (a)>(b)?(b):(a);
  6 inline ll read()//读入优化
  7 {
  8     ll x=0,f=1;//f表示符号,x表示首位数字0
  9     char ch=getchar();
 10     while(ch<'0'||ch>'9')//如果ch不是数字
 11     {
 12         if(ch=='-')//如果是符号就改变符号
 13             f=-1;
 14         ch=getchar();
 15     }
 16     while(ch>='0'&&ch<='9')//如果ch是数字,接下来的每位数字
 17     {
 18         x=x*10+ch-'0';//将数字添加进x内
 19         ch=getchar();
 20     }
 21     return x*f;//返回数值
 22 }
 23 inline void write(ll x)//输出优化
 24 {
 25     if(x<0)//判断小于0的情况
 26     {
 27         putchar('-');
 28         x=-x;
 29     }
 30     if(x>9)//保存每一位
 31     {
 32         write(x/10);
 33     }
 34     putchar(x%10+'0');//输出
 35 }
 36 inline ll gcd(ll a,ll b)
 37 {
 38     return b==0?a:gcd(b,a%b);
 39 }
 40 const ll INF=1ll<<60;
 41 const ll inf=-1ll<<60;
 42 const ll N=4444;
 43 ll pre[N];
 44 bool t[N];//t 用于标记独立块的根结点
 45 inline ll find(ll x)//查找根节点
 46 {
 47     ll r=x;
 48     while(pre[r]!=r)
 49         r=pre[r];//返回根节点 r
 50         int i=x,j;
 51         while(pre[i]!=r)//路径压缩
 52         {
 53             j=pre[i]; // 在改变上级之前用临时变量  j 记录下他的值
 54             pre[i]=r;//把上级改为根节点
 55             i=j;
 56         }
 57     return r;
 58 }
 59 inline void join(ll x,ll y)//判断x y是否连通，
 60 {
 61     ll fx=find(x),fy=find(y);
 62     if(fx!=fy)
 63         pre[fy]=fx;//如果已经连通，就不用管了 //如果不连通，就把它们所在的连通分支合并起来
 64 }
 65 ll n,m;
 66 ll num[N];
 67 int main()
 68 {
 69     while(scanf("%lld%lld",&n,&m)!=EOF)
 70     {
 71         /*
 72         for(i=1;i<=N;i++)
 73             pre[i]=i;//初始化
 74         for(i=1;i<=M;i++)
 75         {
 76             scanf("%d%d",&a,&b);
 77             join(a,b);//判断x y是否连通
 78         }
 79         memset(t,0,sizeof(t));
 80         for(i=1;i<=N;i++)
 81             t[find(i)]=1;//标记根结点
 82         for(ans=0,i=1;i<=N;i++)
 83             if(t[i])
 84             ans++;
 85         printf("%d
",ans-1);
 86         */
 87         for(ll i=1;i<=n;++i)
 88         {
 89             pre[i]=i;
 90         }
 91         memset(num,false,sizeof(num));
 92         ll cnt=0;
 93         for(ll i=0;i<m;++i)
 94         {
 95             ll x,y,z;
 96             x=read();
 97             y=read();
 98             z=read();
 99             if(x==y)
100                 continue;
101             num[y]+=z;
102             num[x]+=z;
103             if(find(x)!=find(y))
104             {
105                 pre[find(x)]=find(y);
106                 ++cnt;
107             }
108         }
109         if(cnt!=n-1)
110         {
111             printf("0
");
112             continue;
113         }
114         ll ans=num[1];
115         for(ll i=2;i<=n;i++)
116             ans=min(ans,num[i]);
117         write(ans);
118         printf("
");
119     }
120     return 0;
121 }

    度度熊与邪恶大魔王

    Accepts: 2114
    Submissions: 13031
    Time Limit: 2000/1000 MS (Java/Others)
    Memory Limit: 32768/32768 K (Java/Others)
    Problem Description

    度度熊为了拯救可爱的公主，于是与邪恶大魔王战斗起来。

    邪恶大魔王的麾下有n个怪兽，每个怪兽有a[i]的生命值，以及b[i]的防御力。

    度度熊一共拥有m种攻击方式，第i种攻击方式，需要消耗k[i]的晶石，造成p[i]点伤害。

    当然，如果度度熊使用第i个技能打在第j个怪兽上面的话，会使得第j个怪兽的生命值减少p[i]-b[j]，当然如果伤害小于防御，那么攻击就不会奏效。

    如果怪兽的生命值降为0或以下，那么怪兽就会被消灭。

    当然每个技能都可以使用无限次。

    请问度度熊最少携带多少晶石，就可以消灭所有的怪兽。

    Input

    本题包含若干组测试数据。

    第一行两个整数n，m,表示有n个怪兽，m种技能。

    接下来n行，每行两个整数，a[i],b[i]，分别表示怪兽的生命值和防御力。

    再接下来m行，每行两个整数k[i]和p[i]，分别表示技能的消耗晶石数目和技能的伤害值。

    数据范围:

    1<=n<=100000

    1<=m<=1000

    1<=a[i]<=1000

    0<=b[i]<=10

    0<=k[i]<=100000

    0<=p[i]<=1000

    Output

    对于每组测试数据，输出最小的晶石消耗数量，如果不能击败所有的怪兽，输出-1

    Sample Input
    1 2
3 5
7 10
6 8
1 2
3 5
10 7
8 6
    Sample Output
    6
18

    题目链接：http://bestcoder.hdu.edu.cn/contests/contest_showproblem.php?cid=774&pid=1003

    分析：完全背包嘛，签到题的说，对着完全背包看看就好咯，不会的请移步这里

     1 #include <bits/stdc++.h>
 2 #include <stdio.h>
 3 using namespace std;
 4 typedef __int64 ll;
 5 #define max(a,b) (a)>(b)?(a):(b);
 6 #define min(a,b) (a)>(b)?(b):(a);
 7 inline ll read()//读入优化
 8 {
 9     ll x=0,f=1;//f表示符号,x表示首位数字0
10     char ch=getchar();
11     while(ch<'0'||ch>'9')//如果ch不是数字
12     {
13         if(ch=='-')//如果是符号就改变符号
14             f=-1;
15         ch=getchar();
16     }
17     while(ch>='0'&&ch<='9')//如果ch是数字,接下来的每位数字
18     {
19         x=x*10+ch-'0';//将数字添加进x内
20         ch=getchar();
21     }
22     return x*f;//返回数值
23 }
24 inline void write(ll x)//输出优化
25 {
26     if(x<0)//判断小于0的情况
27     {
28         putchar('-');
29         x=-x;
30     }
31     if(x>9)//保存每一位
32     {
33         write(x/10);
34     }
35     putchar(x%10+'0');//输出
36 }
37 inline ll gcd(ll a,ll b)
38 {
39     return b==0?a:gcd(b,a%b);
40 }
41 const int N=100010;
42 const ll INF=1ll<<60;
43 const ll inf=-1ll<<60;
44 ll w[N],v[N];
45 ll x[N],y[N];
46 ll dp[20][N];
47 ll n,m;
48 int main()
49 {
50     while(scanf("%I64d%I64d",&n,&m)!=EOF)
51     {
52         for(ll i=1;i<=n;i++)
53             w[i]=read(),v[i]=read();
54         for(ll i=1;i<=m;i++)
55             x[i]=read(),y[i]=read();
56         for(ll i=0;i<=15;i++)
57         {
58             dp[i][0]=0;
59             for(ll j=1;j<=1010;j++)
60                 dp[i][j]=INF;
61             for(ll j=1;j<=m;j++)
62             {
63                 if(y[j]<=i)
64                     continue;
65                 for(ll k=1;k<=1010;k++)
66                 {
67                     ll q=max(k-y[j]+i,0);
68                     dp[i][k]=min(dp[i][k],dp[i][q]+x[j]);
69                 }
70             }
71         }
72         ll ans=0;
73         for(ll i=1;i<=n;i++)
74         {
75             if(dp[v[i]][w[i]]==INF)
76             {
77                 ans=-1;
78                 break;
79             }
80             ans+=dp[v[i]][w[i]];
81         }
82         write(ans);
83         printf("
");
84         //cout<<ans<<endl;
85         //printf("%I64d
",ans);
86     }
87     return 0;
88 }

    度度熊的午饭时光

    Accepts: 375
    Submissions: 5441
    Time Limit: 2000/1000 MS (Java/Others)
    Memory Limit: 32768/32768 K (Java/Others)
    Problem Description

    度度熊最期待每天的午饭时光，因为早饭菜品清淡，晚饭减肥不敢吃太多（胖纸的忧伤T.T）。

    百度食堂的午餐超级丰富，祖国各大菜系应有尽有，度度熊在每个窗口都有爱吃的菜品，而且他还为喜爱的菜品打了分，吃货的情怀呀(>.<)。

    但是，好吃的饭菜总是很贵，每天的午饭预算有限，请帮度度熊算一算，怎样打饭才能买到的最好吃的饭菜?(不超过预算、不重样、午餐等分最高的情况下，选择菜品序号加和最小，加和相等时字典序最小的组合)

    Input

    第一行一个整数T，表示T组数据。 每组测试数据将以如下格式从标准输入读入：

    B

    N

    score_1 cost_1

    score_2 cost_2

    :

    score_N cost_N

    第一行，正整数B（0 <= B <= 1000），代表午餐的预算。

    第二行，正整数N (0 <= N <= 100)，代表午餐可选的菜品数量

    从第三行到第 (N + 2) 行，每行两个正整数，以空格分隔，score_i表示菜品的得分，cost_i表示菜品的价格(0 <= score_i, cost_i <= 100)。

    Output

    对于每组数据，输出两行： 第一行输出："Case #i:"。i代表第i组测试数据。 第二行输出菜品的总得分和总花费，以空格分隔。 第三行输出所选菜品的序号，菜品序号从1开始，以空格分隔。

    Sample Input
    2
29
6
9 10
3 4
6 5
7 20
10 9
15 11
0
2
2 23
10 12
    Sample Output
    Case #1:
34 29
2 3 5 6
Case #2:
0 0

    题目链接：http://bestcoder.hdu.edu.cn/contests/contest_showproblem.php?cid=774&pid=1004

    分析：01背包裸题，要在最大化总得分的情况下最小化序号之和，并输出字典序最小的解，在不打饭的情况下不输出空行，嗯，就是介个样子！

    下面给出AC代码：

      1 #include <stdio.h>
  2 #include <string.h>
  3 #include <bits/stdc++.h>
  4 using namespace std;
  5 typedef long long ll;
  6 #define max(a,b) (a)>(b)?(a):(b);
  7 #define min(a,b) (a)>(b)?(b):(a);
  8 inline ll read()//读入优化
  9 {
 10     ll x=0,f=1;//f表示符号,x表示首位数字0
 11     char ch=getchar();
 12     while(ch<'0'||ch>'9')//如果ch不是数字
 13     {
 14         if(ch=='-')//如果是符号就改变符号
 15             f=-1;
 16         ch=getchar();
 17     }
 18     while(ch>='0'&&ch<='9')//如果ch是数字,接下来的每位数字
 19     {
 20         x=x*10+ch-'0';//将数字添加进x内
 21         ch=getchar();
 22     }
 23     return x*f;//返回数值
 24 }
 25 inline void write(ll x)//输出优化
 26 {
 27     if(x<0)//判断小于0的情况
 28     {
 29         putchar('-');
 30         x=-x;
 31     }
 32     if(x>9)//保存每一位
 33     {
 34         write(x/10);
 35     }
 36     putchar(x%10+'0');//输出
 37 }
 38 inline ll gcd(ll a,ll b)
 39 {
 40     return b==0?a:gcd(b,a%b);
 41 }
 42 const ll INF=1ll<<60;
 43 const ll inf=-1ll<<60;
 44 const ll N=1510;
 45 ll w[N],v[N];
 46 ll u[N];
 47 ll dp[N][N];
 48 bool vis[N][N];
 49 //-------------------------------------------
 50 inline ll solve(ll x,ll y)
 51 {
 52     ll pp=dp[x][0];
 53     ll qq=dp[y][0];
 54     for(ll p=1,q=1;p<=pp&&q<=qq;++p,++q)
 55     {
 56         if(dp[pp][p]!=dp[qq][q])
 57             return dp[pp][p]-dp[qq][q];
 58     }
 59     return 0;
 60 }
 61 //-------------------------------------------
 62 ll T,n,m;
 63 int main()
 64 {
 65     while(scanf("%lld",&T)!=EOF)
 66     {
 67     //T=read();
 68         for(ll i=1;i<=T;++i)
 69         {
 70             m=read();
 71             n=read();
 72             memset(u,false,sizeof(u));
 73             memset(vis,false,sizeof(vis));
 74             memset(dp,false,sizeof(dp));
 75             memset(v,false,sizeof(v));
 76             memset(w,false,sizeof(w));
 77             for(ll j=1;j<=n;++j)
 78             {
 79                 v[j]=read();
 80                 w[j]=read();
 81             }
 82             printf("Case #%lld:
",i);
 83             for(ll j=1;j<=n;++j)
 84             {
 85                 for(ll k=m;k>=w[j];--k)
 86                 {
 87                     //ll a=u[k];
 88                     //u[k]=max(u[k],u[k-w[j]]+v[j]);
 89                     //if(a!=u[k])
 90                        //vis[j][k]=1;
 91                     if(u[k]<u[k-w[j]]+v[j])
 92                     {
 93                         u[k]=u[k-w[j]]+v[j];
 94                         vis[j][k]=true;
 95                     }
 96                 }
 97             }
 98             //
 99             ll maxn=0;
100             for(ll j=0;j<=m;++j)
101                 maxn=max(maxn,u[j]);
102             //
103             ll sum=INF;
104             ll x=0,num=0;
105             for(ll j=m;j>=0;--j)
106             {
107                 if(u[j]==maxn)
108                 {
109                     ll sumsum=0;
110                     ll pre=1;
111                     ll xx=n,yy=j;
112                     while(xx>=1&&yy>=0)
113                     {
114                         if(vis[xx][yy])
115                         {
116                             dp[x][pre++]=xx;
117                             //pre++;
118                             sumsum+=xx;
119                             yy-=w[xx];
120                         }
121                         xx--;
122                     }
123                     dp[x][0]=pre-1;
124                     sort(dp[x]+1,dp[x]+1+dp[x][0]);
125                     if(sum>sumsum)
126                     {
127                         sum=sumsum;
128                         num=x;
129                     }
130                     else if(sum==sumsum&&solve(x,num)<0)
131                     {
132                         sum=sumsum;
133                         num=x;
134                     }
135                     x++;
136                 }
137             }
138             ///
139             ll val=0,cost=0;
140             ll top=dp[num][0];
141             sort(dp[num]+1,dp[num]+1+dp[num][0]);
142             for(ll a=1;a<=top;++a)
143             {
144                 val+=v[dp[num][a]];
145                 cost+=w[dp[num][a]];
146             }
147             //printf("Case #%lld:
",i);
148             printf("%lld %lld
",val,cost);
149             for(ll a=1;a<top;++a)
150                 printf("%lld ",dp[num][a]);
151             if(top>=1)
152                 printf("%lld
",dp[num][top]);
153         }
154     }
155     return 0;
156 }

    寻找母串

    Accepts: 82
    Submissions: 676
    Time Limit: 2000/1000 MS (Java/Others)
    Memory Limit: 32768/32768 K (Java/Others)
    Problem Description

    对于一个串S，当它同时满足如下条件时，它就是一个01偏串： 1、只由0和1两种符组成； 2、在S的每一个前缀中，0的个数不超过1的个数； 3、S中0的个数和1的个数相等。

    现在给定01偏串S，请计算一下S在所有长度为n的01偏串中作为子串出现的次数的总和。 由于结果比较大，结果对1e9+7取余后输出。

    样例解释： 在第二个样例中，长度为4的偏串共两个1010，1100。10在1010中出现了两次，在1100中出现了1次。所以答案是3。

    Input

    第一行给出一个整数T(1<=T<=40),表示测试数据的数目。 每一组测试包含一个整数n和字符串S，中间用空格分开。(1<=|S|<=100000，1<=n<=1000000000)

    输入保证S是一个01偏串。

    Output

    对于每一组数据，输出一个整数占一行，表示答案。

    Sample Input
    2
2 10
4 10
    Sample Output
    1
3

    题目链接：http://bestcoder.hdu.edu.cn/contests/contest_showproblem.php?cid=774&pid=1005

    分析：

    

    这题看起来束手无策，但其实你可以先写个暴力观察小数据，这个时候你就会发现其实这题比你想象的要简单的多
    

    

    （比如你会发现其实答案之和字符串的长度|S|有关，和字符串的内容是无关的）
    

    

    没错！这题有规律，答案就是
    

    


    

    

    

    但是别高兴的太早，这个时候你又会发现其实这题比你想象的要难得多
    

    

    因为这题的n范围巨大（约10亿）而求10亿的组合数是做不到的
    

    

    所有先考虑化简公式看看，令x = (n-|S|)/2+1有
    

    


    

    

    

    其中F[x]是第x项卡特兰数
    


    通项公式：F[n] = C(2n, n)/(n+1) = C(2n, n)-C(2n, n-1)
    


    递推公式：F[n+1] = 2*(2*n+1)/(n+2)*F[n]
    


    F[n] = F[0]*F[n-1]+F[1]*F[n-2]+…+F[n-1]*F[0]
    


    

    而卡特兰数有O(n)的递推公式
    

    


    

    

    

    这样的话理论上可以O(n)求出所有的答案，可还是不行。。。
    

    

    复杂度已经不能再优化了。。所以只能考虑打表
    

    

    可是你又开不了10亿的数组，但是这题也没有说要O(1)询问呀
    

    

    没错！分块打表
    

    

    你只需要后台O(n)暴力出第100000个卡特兰数，第200000个卡特兰数……第500000000个卡特兰数就好了
    

    

    也就是开个5000+的数组s[]，其中s[i]是第100000*i个卡特兰数
    

    

    然后对于每组询问(n, |S|)，先计算x=(n-|S|)/2+1，然后再找到x所在的那一块（比x小离x最近的100000的倍数），
    

    

    之后暴力转移就好，最多转移100000次
    

    

    因为有除法所以要乘法逆元，总体复杂度O(100000log(n))
    

    

    还有注意n为奇数的时候答案一定为0，因为母串要保证0和1的数量相等，所以不存在长度为奇数的母串
    

    

    除此之外，n<|S|答案也为0
    


    接下来只以10101010101010为例
    模式串：                 字串：                               匹配条件：
    10                      10                                  10              1        total        1
    1010                    10                                  1010            2        
                                                                1100            1        total        3
    101010                  10                                  101010          3
                                                                101100          2
                                                                110100          2
                                                                110010          2
                                                                111000          1        total         10
    10101010                10                                  10101010        4
                                                                10101100            3
                                                                                 10110100        3
                                                                11010100        3
                                                                10110010        3
                                                                11001010        3
                                                                11010010        3
                                                                11011000        2
                                                                11001100        2
                                                                10111000        2
                                                                11101000        2
                                                                11100100        2
                                                                11100010        2
                                                                11110000        1         total        35

    即组合数C（n*2-1，n）n=（n-strlen（s））/2+1；
    =f[n]*(n+1)/2;
    f[n]=2*(2*n-1)*f[n-1]/(i+1);
    下面给出AC打表代码：(打了一天的表)
    

    

      1 #include <bits/stdc++.h>
  2 using namespace std;
  3 typedef long long ll;
  4 const ll N=1000010;
  5 #define max(a,b) (a)>(b)?(a):(b);
  6 #define min(a,b) (a)>(b)?(b):(a);
  7 inline ll read()//读入优化
  8 {
  9     ll x=0,f=1;//f表示符号,x表示首位数字0
 10     char ch=getchar();
 11     while(ch<'0'||ch>'9')//如果ch不是数字
 12     {
 13         if(ch=='-')//如果是符号就改变符号
 14             f=-1;
 15         ch=getchar();
 16     }
 17     while(ch>='0'&&ch<='9')//如果ch是数字,接下来的每位数字
 18     {
 19         x=x*10+ch-'0';//将数字添加进x内
 20         ch=getchar();
 21     }
 22     return x*f;//返回数值
 23 }
 24 inline void write(ll x)//输出优化
 25 {
 26     if(x<0)//判断小于0的情况
 27     {
 28         putchar('-');
 29         x=-x;
 30     }
 31     if(x>9)//保存每一位
 32     {
 33         write(x/10);
 34     }
 35     putchar(x%10+'0');//输出
 36 }
 37 ll pre[N];
 38 bool t[N];//t 用于标记独立块的根结点
 39 inline ll find(ll x)//查找根节点
 40 {
 41     ll r=x;
 42     while(pre[r]!=r)
 43         r=pre[r];//返回根节点 r
 44         int i=x,j;
 45         while(pre[i]!=r)//路径压缩
 46         {
 47             j=pre[i]; // 在改变上级之前用临时变量  j 记录下他的值
 48             pre[i]=r;//把上级改为根节点
 49             i=j;
 50         }
 51     return r;
 52 }
 53 inline void join(ll x,ll y)//判断x y是否连通，
 54 {
 55     ll fx=find(x),fy=find(y);
 56     if(fx!=fy)
 57         pre[fy]=fx;//如果已经连通，就不用管了 //如果不连通，就把它们所在的连通分支合并起来
 58 }
 59 const ll mod=1000000007;
 60 inline ll gcd(ll a,ll b)
 61 {
 62     return b==0?a:gcd(b,a%b);
 63 }
 64 inline ll qpow(ll x,ll p)
 65 {
 66     ll ret=1;
 67     for(;p;p>>=1,x=x*x%mod)
 68     {
 69         if(p&1)
 70         ret=ret*x%mod;
 71     }
 72     return ret;
 73 }
 74 const ll INF=1ll<<60;
 75 const ll inf=-1ll<<60;
 76 char str[N];
 77 ll s[N]=
 78 {
 79     1,945729344,874271714,728937729,488957151,213941567,621948608,165410919,130152966,477849061,70646122,104331417,804334677,112228257,406385851,373610584,959189339,129232103,596011761,653302265,502350404,254254274,23933464,200023312,437339905,838455202,353529507,181826790,268707842,918554769,361346624,475114772,551764204,264122255,348587014,339775734,190610810,147039338,576212642,958452888,907080272,897468755,485757524,926001334,253159604,720502565,795701677,485379561,512490195,62869610,44647373,170729326,181183678,223602264,14266485,546070015,669748641,266401585,722432477,844276843,806809752,215407850,514089950,48258832,388402233,287386547,184275733,39087220,221157750,796058916,274451201,551481903,468840632,654694122,349643823,123450237,466554578,613371117,104229634,534520804,340376885,53768135,665006362,362834988,113383960,458629513,877935697,990187981,24010351,477599350,491321663,619368665,730486899,302800222,248364522,617385859,136706557,988949479,681384128,140672609,39443745,327846413,353172956,108275948,378351259,482275913,432213793,904945283,871063203,757590126,541675614,649654378,339103071,153511190,307061975,535998427,953278988,315396128,629843192,166102135,623766767,9922024,962639966,258645380,484156059,75913669,837794293,574075634,404396328,768092500,903633183,512519672,44285819,538902069,621789723,71341556,81438997,28847533,40513591,849551327,716939958,942552133,395636757,409774346,383255288,612104482,95851638,696370843,884779483,524922348,220269701,390881347,114227221,395767320,614338019,380172060,180822541,691300228,891829295,967678016,701386253,162940737,994175698,908221589,19098271,438672588,631720803,573850469,339056525,473356393,859071294,949727021,98529638,168890563,563975847,7483795,276476730,39489314,514728211,87529508,963753157,380031470,856696972,158703490,515908206,227044626,959511111,993078622,80360936,888779384,323285665,543535546,550763954,601069950,458756853,717452438,141044757,197501322,634124805,238583278,200414323,553793080,66504653,42069842,722598090,913608032,740683059,742050754,961341417,612741304,91785690,526087340,214299335,290988315,920349511,650032143,826861662,655676777,832872230,243503280,428075391,780176613,762279490,958553222,298583588,683539087,245173409,561149407,172900450,994863083,373712992,577710034,676813275,65061172,354158163,474587452,779048700,252731521,513247255,63716789,589280106,300449755,843219077,437936157,617131692,778975010,166446598,341293627,939851882,377019792,974135048,18159888,617899054,824539871,922256819,904951680,298436189,877366713,763013330,69488053,73161945,968841940,502298660,815060335,585478837,52225572,617841877,964952304,864682249,537138023,438032312,507344261,611203109,211965075,510592131,253042462,662211654,306668985,993054101,886017682,550897294,748379779,600811784,662510825,923914254,398819263,459843924,563578785,800110936,890507874,833696238,67588077,662523469,355339772,897895240,326276953,66185485,707119691,543399352,359252814,343020869,435649016,326479513,313711009,180736295,933688638,341979700,832484396,852821336,938966456,327938691,897898594,514258765,67631305,832209634,147167201,933864871,764372915,290610995,894446445,552066969,955550128,899538317,98587581,791520663,371485442,250166899,386277652,850700239,24883422,365069578,542759096,393648297,338311897,998392022,963038213,184331347,216025562,616142588,243671535,922469047,642896156,173840546,617065897,502916270,399859525,626708892,203438585,794335044,466847771,485435744,176762177,370414892,387922877,421040690,506803021,831791096,536761402,659624740,205742555,354512880,992080290,655638247,441557017,45234385,602453687,642559130,907907516,150309994,959349261,667245533,336984599,7514180,293021893,527065808,716061908,183191397,64976751,577967909,833760679,814950379,238991246,714290102,902231891,437427307,149073027,103173270,685925477,694198173,725256565,71168984,261760780,55757530,112359206,436686978,516723667,537954237,916255072,34043384,456908325,881291767,146092388,189805926,498850664,471576564,977024064,502710283,230937450,166248501,100683033,725464484,750575586,525293805,107750423,541103694,100780834,184498203,767490572,317419646,113029816,182757677,138614676,694365799,832641460,450632411,301670090,972578606,809851941,356139480,741575534,319109663,491688831,946837771,820876500,944731056,441691410,439779726,878100434,110259934,781810231,822737100,122949140,185157201,147977057,78103215,98159640,570747872,913214189,526507036,614910594,606219102,234699129,882201646,184318349,623751699,220799331,756153289,390599249,416019375,679472812,608212465,946818526,676784388,848462543,701625645,631586665,279738981,507735946,
 80     983863013,797150469,667058957,302361878,11123699,265956565,759232458,630434107,603234456,651468455,184824418,604726700,101186449,24954474,873710205,172883575,299713548,965606121,589841373,782619807,545740664,94087806,846145323,99057958,275906334,98324641,712417499,506415720,153372884,988856645,995466415,137196894,296743168,461577383,447394859,350928543,800328411,359672375,623997509,289025167,258422311,586168764,714133593,295537278,650382172,140783549,886171581,956390396,255289567,255216580,795608020,678371829,996937306,387496328,235100473,649738138,898466097,407604566,529571196,854241604,278302571,998505802,446819056,674284500,819876855,166920818,141076505,198714798,612486167,700718656,391542764,645522969,999548402,477521930,505594128,951413460,665706996,664760847,514065631,101412856,896853242,846607158,200777626,574295366,319187233,35082951,154276754,661540946,477502005,725521461,655895869,355739306,678518235,192498927,173855439,165487057,456827891,29739781,258791510,428391536,531278508,763339063,168419466,868466779,365442496,722700915,698024831,661694896,590895521,463918561,314021483,842517461,302478861,688035954,232366905,961113387,828397927,320767088,508824951,735431855,426624942,591154618,76109991,328101502,665849594,385026767,472223460,891042862,342446307,246096980,406346591,527916773,998217449,259406947,423766371,70821164,405777406,55383694,177073358,520289379,953144123,598695706,882910945,148088957,127593638,565277379,746539154,329767227,652704430,458041013,532152689,943689273,198275398,746095184,296234153,567962117,585755152,914398975,392347020,605551281,526603889,299409566,554048889,207398267,231395425,761907727,542580836,511879608,477430193,68337675,420520000,90537136,881957003,293703031,601859481,982488742,687784417,778652850,678832519,41785321,569973809,659662535,165782945,800601868,946694094,798296737,530957874,155643917,78799254,840622486,223970795,8485824,28964502,945323929,340741737,34732972,603272702,607664609,673843887,72927622,72008879,956047483,777700919,664146374,919751340,24878865,460990642,148315028,357424860,741503968,317099797,829104947,566163262,167156841,600465821,697110956,353280563,96899824,252662837,7893272,120921870,514991140,889565398,248085788,692098208,933756018,425884820,645876939,21896522,887680365,852291368,441176290,66262852,156844282,636816511,124116038,842190463,966404243,788403295,940665695,989549719,828820601,250027891,298842711,270122615,174406628,2135753,46964343,877722285,279439798,165618275,192058083,45442983,312364959,141906913,672563513,989170139,871761564,190705367,121151650,442674415,781847041,760092753,473161041,775211615,133246058,75660748,105103896,314024082,699785781,135093763,341755252,356922254,545665155,516050961,683432858,554252611,980150710,3809442,280205936,618138190,143338978,788688497,335020131,301878323,399780697,432488867,731503292,78552686,970031600,634518684,501384623,44760617,805930887,940017151,197445475,720089820,203297800,544488568,323817708,248584685,735890324,467669218,62087980,905110436,891634369,10533152,23990287,611609572,542849349,924481418,4214140,12444440,736717687,401040098,157325666,367265798,451303368,161946117,507226623,459380345,866752949,897212880,806566311,378612676,64109037,642588978,48895787,519942317,136434876,29861018,573535950,758513856,146511954,726260224,399742173,745185429,190644678,862659605,284303510,93481167,54655028,557236856,907886262,184073171,469064478,587825128,561305691,650561892,494974250,427901529,127482682,377916374,510109927,601476427,61867544,242080407,336520612,817284238,357107802,831433550,401060931,360517611,209089633,445727073,273046955,399724590,819481328,852463455,937507966,711938640,599907247,439538485,872938446,909790292,180096711,650114586,529418274,560890330,882019804,509410348,501169449,344529911,730231156,858736987,621110787,4459667,873347202,696188165,506145206,73642136,147943559,379421898,654783213,956981417,651743441,639780318,78577382,593374499,830162763,301755654,792435305,117970707,932913412,832255341,132217712,628450721,227843471,107653574,884405265,36919433,120185702,713929739,355335147,939617354,328252980,400671521,3750430,291712420,161768446,204525632,110812455,655316736,810170496,495246572,411585803,110977744,480012790,837775503,421678299,537576259,215096702,255249918,999498556,551932245,400608941,789747750,660573311,224129758,960267908,823407521,663185418,71834349,194653367,479377667,556838543,30135513,434301805,554957759,766164228,423757881,77024851,446918481,86585877,167204113,199048610,165982142,29904724,116697346,136404554,840462862,989156366,172584177,788447557,850533198,955155103,112270248,718115841,395066112,581133650,
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 83     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 84     613098019,82701766,247051132,940988662,699855854,653160599,253589170,454574426,648632475,561229480,439013878,780463812,751080642,475363127,872955171,221419431,660718935,616253276,620991239,322955474,95900697,736953117,435364733,400032675,143150458,892976596,778936048,870114869,693427846,916920202,531290913,159313409,854027823,956895320,431001333,958032952,435621676,147281691,782644049,363417645,788629073,318986652,45280716,627497306,862125833,55176692,769605990,50913091,759547794,415399356,91873313,435971457,972361711,886131838,698360174,778537252,867612407,363143588,210628827,227452019,167347791,136481555,794414626,886837851,833212071,867100102,962413595,506550918,150461107,97396199,322574611,286566460,418048639,126619502,49890932,424873463,577636888,598651314,214771282,633007716,67600639,872694032,899190404,807799524,536389663,783650145,198103723,406176925,596091368,865801146,255660985,950263612,204698161,570789483,35674858,355559169,627978911,108585701,612463851,227550456,44549978,430769857,197325621,365494994,215296621,794175066,165486775,239680547,990696203,54536891,267132823,879937767,468746039,925979058,896931201,784849771,365504305,221529004,715124186,598394296,273404498,942530408,892470275,339529469,637182412,714030555,4591038,747727171,993858038,98351859,587568276,635308938,66381702,523069101,611647432,93089964,621299142,87170109,91184576,582205309,384761096,623784317,274148172,565367846,707139350,413965784,111585803,462251445,228866268,199084146,547319848,912054783,776081313,50512733,398666776,860515976,132645483,784732014,283636962,567053755,762371460,902545259,6795752,933181225,79095380,382697365,79944172,18195218,904541583,983398375,352534501,206544638,590062152,870793758,26235371,885333891,755593357,797697319,233605783,916638079,592465092,389607454,700420311,196459900,86420454,160839557,696575,569324134,807071644,759233878,743587997,984681992,941063338,324982757,676014860,600520498,188292161,129801562,413126752,313832643,937070759,97526604,541466777,603982189,77969188,347408805,311339522,716968535,967397279,253173933,37355703,352327793,883519616,967973262,117140785,240224335,807324871,845415681,840729498,935684923,155577573,734434833,191517923,654123163,637896624,852058670,173168370,730184794,50915735,324841114,628137110,205149114,182695122,548334792,230832142,246897375,673812822,774458235,12653594,589139188,823592104,141584849,661263989,634220976,838048410,865644208,959538649,496526723,706515642,30364288,288996675,618298630,512675413,255909709,961226649,196140093,464842799,73771491,907822468,820941783,633421126,938423057,725611016,889132555,244245553,600535214,454683334,918501588,156397960,727549083,462344797,552581171,951585647,331017637,713844297,201014346,220893764,942832265,205089583,855055766,558242708,467800323,544040124,5018684,600324086,275710404,942082597,882387855,283394592,984698042,699687145,218659340,671476043,786242938,615660310,451633451,264041371,62057607,615318684,358229710,600528374,737949511,309417464,928413938,485704507,508289212,335423875,674380863,518017044,336119074,617416674,485819110,941096637,929311054,597002446,973475334,265580872,563517784,251569149,148530184,95396816,883600979,282561244,840871566,519908445,640741041,421822961,632376045,139437876,790411904,591820303,741935921,948651905,366299783,716525381,15431695,15465625,971346451,168728596,608506519,27394837,851151360,602776233,970664278,322046939,511916124,340275737,1956214,363111458,481072216,658163046,518552634,132664065,981823198,216243081,159795020,767941105,457986980,858463537,976190745,171187323,572871813,647899578,408338249,44147207,156374207,940179853,58043446,28385650,424517633,511617049,926045094,319879729,514748429,614317869,421026423,671332980,6835612,938930076,458303594,925057217,380172144,417162197,8078854,424003560,2008065,807569105,429700641,464740805,457368858,589073747,394260986,391506131,929826825,432872648,690774587,357406694,692130031,273616783,445027156,511488576,31976171,170358186,20482390,918198464,271438684,749710986,389073776,939445726,23455846,956187346,748137028,734748051,303817400,94204386,126522637,81942995,832809073,197225686,406203721,149560387,711052519,501674175,547993890,871374386,551852029,842913905,11301805,658024191,322045818,938590456,235068927,986201888,729075821,230293940,88489051,430417352,457640653,512463037,164939381,78855727,918299124,19019884,318003658,964003586,887976673,465062825,998287188,345218097,418436509,423868376,737852682,222443100,431664080,191790675,179137865,240191988,426955600,372855698,834981397,769820197,12457840,592849249,301573725,529831284,455945822,908361093,348036200,725019234,272002918,302014486,868309988,267131588,110362906,658933926,714755636,575129133,572255614,242451399,460854477,379734453,261883330,173252303,205271891,875823576,295295415,344691305,88281196,291229175,339458347,199662270,135955543,945346312,464533492,871840955,247580453,899980261,309580183,86978522,845735069,219736536,817617603,405963484,880265007,291366034,314006019,86753505,849126155,994666022,724217006,569965800,190010599,920028225,755930223,376881817,18255664,401402435,819561952,810240409,896675374,5648890,14322856,899043492,141273287,496383310,791004873,659659128,952106154,219105831,176538445,879815182,445593480,150528189,94493662,797589952,732282876,567925839,796782054,629643845,130694449,475573885,497611040,499200426,183468775,892394136,17028961,825354772,382036602,10367014,453970224,702396666,603040526,616353655,902691380,983386965,692566149,247825935,66330622,225112465,895795849,52320773,155947313,718224149,164300545,677709978,266340441,218514417,204476400,286595553,643512303,598766962,788301683,892598856,520036990,859133788,979526775,296869156,670015326,164319827,647208607,679949247,172064170,865163183,337039103,105046080,579964945,123270759,385113924,576031432,771688525,704910535,414466742,609386035,928631412,485368568,479480031,505949447,116422341,568991478,93574014,961174363,735174682,235632013,19234330,650946685,144392601,510837538,567469814,856042956,806198989,116886696,207340279,223268817,253945420,584643811,616483622,711150993,357748602,626063923,607222114,600586776,28542569,796083795,549421818,591429425,607254162,355391295,492527448,847958740,287070353,814226472,643144229,698819990,177700395,611101671,490661583,275559707,950224920,844587825,53229930,339523379,979109376,126378471,444745387,726629825,351823753,475484131,518938465,963071536,84665553,643867081,825793617,408943841,879707301,471828659,223598556,384354814,153231894,188992442,483627866,255018189,675911377,500327547,638084035,486199656,903986764,949245648,862319148,790947898,142578126
 85 };
 86 ll T,n;
 87 int main()
 88 {
 89     ios::sync_with_stdio(false);
 90     T=read();
 91     while(T--)
 92     {
 93         scanf("%d%s",&n,&str);
 94         ll len=strlen(str);
 95         if(len>n||n&1)
 96         {
 97             printf("0
");
 98             continue;
 99         }
100         ll pp=(n-len)/2+1;
101         ll x=pp/100000;
102         ll ans=s[x];
103         for(ll i=x*100000+1;i<=pp;++i)
104             ans=(ans*2*(i*2-1)%mod*qpow(i+1,mod-2)%mod)%mod;
105         ans=(ans*(pp+1)%mod*qpow(2,mod-2)%mod)%mod;
106         write(ans);
107         printf("
");
108     }
109     return 0;
110 }
    

    

    


     
    


     
    

    
    
          
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            原文地址：https://www.cnblogs.com/ECJTUACM-873284962/p/7291748.html
          
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