Codeforces 474(#271 Div 2) 解题报告

A：直接贴过来装换

解题代码

 1 // File Name: a.cpp
 2 // Author: darkdream
 3 // Created Time: 2014年10月06日 星期一 23时28分57秒
 4 
 5 #include<vector>
 6 #include<list>
 7 #include<map>
 8 #include<set>
 9 #include<deque>
10 #include<stack>
11 #include<bitset>
12 #include<algorithm>
13 #include<functional>
14 #include<numeric>
15 #include<utility>
16 #include<sstream>
17 #include<iostream>
18 #include<iomanip>
19 #include<cstdio>
20 #include<cmath>
21 #include<cstdlib>
22 #include<cstring>
23 #include<ctime>
24 #define LL long long
25 
26 using namespace std;
27 char  str[102];
28 char  tp[4][100] = {"qwertyuiop",
29 "asdfghjkl;",
30 "zxcvbnm,./",
31 };
32 int main(){
33     scanf("%s",str);
34     if(str[0] == 'L')
35     {
36        scanf("%s",str);
37        for(int i = 0;i < strlen(str);i ++)
38        {
39             for(int j = 0;j <= 2;j ++)
40             {
41                if(strchr(tp[j],str[i]))
42                {
43                  printf("%c",tp[j][strchr(tp[j],str[i]) - tp[j] +1]);
44                }
45             }
46        }
47     }
48   else{
49        scanf("%s",str);
50        for(int i = 0;i < strlen(str);i ++)
51        {
52             for(int j = 0;j <= 2;j ++)
53             {
54                if(strchr(tp[j],str[i]))
55                {
56                  printf("%c",tp[j][strchr(tp[j],str[i]) - tp[j] - 1]);
57                }
58             }
59        
60        }
61   }
62 return 0;
63 }

View Code

B：一个标记或者二分就行，最大1e6的复杂度

 1 // File Name: b.cpp
 2 // Author: darkdream
 3 // Created Time: 2014年10月06日 星期一 23时39分31秒
 4 
 5 #include<vector>
 6 #include<list>
 7 #include<map>
 8 #include<set>
 9 #include<deque>
10 #include<stack>
11 #include<bitset>
12 #include<algorithm>
13 #include<functional>
14 #include<numeric>
15 #include<utility>
16 #include<sstream>
17 #include<iostream>
18 #include<iomanip>
19 #include<cstdio>
20 #include<cmath>
21 #include<cstdlib>
22 #include<cstring>
23 #include<ctime>
24 #define LL long long
25 
26 using namespace std;
27 int a[100005];
28 int b[100005];
29 int n , m ; 
30 int find(int x)
31 {
32     int l = 1; 
33     int r = n;
34     while(l <= r)
35     {
36       int mid = (l + r)/2;
37       if(a[mid] < x)
38       {
39         l = mid + 1;
40       }else {
41         r = mid - 1;
42       }
43     }
44     return l;
45 }
46 int main(){
47   scanf("%d",&n);
48   for(int i =1 ;i <= n;i ++)
49   {      scanf("%d",&a[i]);
50       a[i] = a[i-1] + a[i];
51   }
52   scanf("%d",&m);
53   for(int i =1 ;i <= m;i ++)
54   {
55       scanf("%d",&b[i]);
56       printf("%d
",find(b[i]));
57   }
58 return 0;
59 }

View Code

C：知道转换公式以后就是一个手速题

  1 // File Name: c.cpp
  2 // Author: darkdream
  3 // Created Time: 2014年10月07日 星期二 00时41分28秒
  4 
  5 #include<vector>
  6 #include<list>
  7 #include<map>
  8 #include<set>
  9 #include<deque>
 10 #include<stack>
 11 #include<bitset>
 12 #include<algorithm>
 13 #include<functional>
 14 #include<numeric>
 15 #include<utility>
 16 #include<sstream>
 17 #include<iostream>
 18 #include<iomanip>
 19 #include<cstdio>
 20 #include<cmath>
 21 #include<cstdlib>
 22 #include<cstring>
 23 #include<ctime>
 24 #define LL long long
 25 #define eps 1e-8
 26 using namespace std;
 27 struct node{
 28     double x,y,a,b;
 29 };
 30 node a[5][5];
 31 node  change(node tp)
 32 {
 33    node ans;
 34    ans.a = tp.a;
 35    ans.b = tp.b;
 36    ans.x = tp.a - (tp.y - tp.b);
 37    ans.y = tp.b + (tp.x - tp.a);
 38    return ans;
 39 }
 40 int ans[0];
 41 double dis[5][5];
 42 double thedis(double x1, double y1, double x2,double y2)
 43 {
 44   return sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2)) ;
 45 }
 46 int mi; 
 47 bool ok()
 48 {
 49    node tt[5];
 50    for(int i = 1;i <= 4;i ++)
 51    {
 52       tt[i] = a[i][ans[i]];
 53    }
 54    for(int i = 1;i <= 4;i ++)
 55    {
 56        for(int j= 1;j <= 4 ; j ++)
 57        {
 58             dis[i][j] = thedis(tt[i].x,tt[i].y,tt[j].x,tt[j].y);
 59        }
 60      sort(dis[i] + 1,dis[i] + 1 + 4);
 61    }
 62    for(int i = 2;i <= 4;i ++)
 63        for(int j = 1;j <= 4; j ++)
 64        {
 65           if(fabs(dis[i][j] - dis[i-1][j]) > eps)
 66               return 0 ; 
 67        }
 68    if(fabs(dis[1][4]-0 ) <eps)
 69        return 0 ; 
 70    if(fabs(dis[1][3] - dis[1][2]) > eps)
 71        return 0 ;
 72    if(fabs(dis[1][3] * sqrt(2) - dis[1][4]) > eps)
 73        return 0 ;
 74    if(fabs(dis[1][3] - 0 ) < eps)
 75        return 0 ;
 76    if(fabs(dis[1][2] - 0 ) < eps)
 77        return 0 ;
 78    /*for(int i = 1;i <= 4;i ++)
 79    {
 80        for(int j = 1;j <= 4; j ++)
 81            printf("%lf ",dis[i][j]);
 82        printf("
");
 83    }
 84    for(int i = 1;i<= 4;i ++)
 85        printf("%lf %lf
",tt[i].x,tt[i].y);
 86    
 87        */
 88    return 1;
 89 }
 90 void dfs(int k)
 91 {
 92    if(k == 5)
 93    {
 94       if(ok())
 95       {
 96         if(ans[1] + ans[2] + ans[3] + ans[4]  - 4< mi)
 97            mi = ans[1] + ans[2] + ans[3] + ans[4] -4;
 98         //printf("%d %d %d %d
",ans[1],ans[2],ans[3],ans[4]);
 99       }
100       return;
101    }
102    for(int i = 1;i <= 4;i ++)
103    {
104       ans[k] = i ;
105       dfs(k+1);
106    }
107 }
108 int main(){
109    int n; 
110    scanf("%d",&n);
111    while(n--)
112    {
113        for(int i= 1;i <= 4;i ++)
114           scanf("%lf %lf %lf %lf",&a[i][1].x,&a[i][1].y,&a[i][1].a,&a[i][1].b);
115        for(int i = 1;i<= 4;i ++ ) 
116        {
117           for(int j = 2;j <= 4;j ++)
118               a[i][j] = change(a[i][j-1]);
119        }
120       /* for(int i = 1;i<= 4;i ++ ) 
121        {
122           for(int j = 1;j <= 4;j ++)
123               printf("%lf %lf ** ",a[i][j].x,a[i][j].y);
124           printf("
");
125        }*/
126       mi = 1000; 
127       dfs(1);
128       if(mi == 1000)
129           printf("-1
");
130       else printf("%d
",mi);
131    }
132 return 0;
133 }

View Code

D： dp，考虑最后这一位是R还是W即可

 1 // File Name: d.cpp
 2 // Author: darkdream
 3 // Created Time: 2014年10月07日 星期二 00时21分01秒
 4 
 5 #include<vector>
 6 #include<list>
 7 #include<map>
 8 #include<set>
 9 #include<deque>
10 #include<stack>
11 #include<bitset>
12 #include<algorithm>
13 #include<functional>
14 #include<numeric>
15 #include<utility>
16 #include<sstream>
17 #include<iostream>
18 #include<iomanip>
19 #include<cstdio>
20 #include<cmath>
21 #include<cstdlib>
22 #include<cstring>
23 #include<ctime>
24 #define LL long long
25 #define maxn 100004
26 #define M 1000000007
27 using namespace std;
28 int t, k ;
29 
30 LL ans[maxn][2];
31 LL sum[maxn];
32 int main(){
33   while(scanf("%d %d",&t,&k) != EOF)
34   {
35      ans[1][1] = 1;
36      ans[0][2] = 1; 
37      
38      if(k == 1 )
39        ans[1][2] = 1;
40      else ans[1][2] = 0 ; 
41      sum[0] = 0 ;
42      sum[1] = ans[1][1] + ans[1][2];
43      for(int i = 2; i <= 100000;i ++)
44      {
45          ans[i][1] = (ans[i-1][1] + ans[i-1][2])%M ;
46          if(i >= k )
47            ans[i][2] = (ans[i-k][1] + ans[i-k][2])%M;
48          else{
49            ans[i][2] = 0 ; 
50          }
51          sum[i] = (sum[i-1] + ans[i][1] + ans[i][2])%M;
52     //    printf("%I64d
",sum[i]);
53      }
54      int a, b; 
55      for(int i =1 ;i <= t;i ++)
56      {
57        scanf("%d %d",&a,&b);
58        printf("%I64d
",((sum[b] - sum[a-1])+ M)%M );
59      }
60      
61   }
62 return 0;
63 }

View Code

E：离散化，二分找端点值，线段树查找最大值

  1 // File Name: e.cpp
  2 // Author: darkdream
  3 // Created Time: 2014年10月07日 星期二 01时29分07秒
  4 
  5 #include<vector>
  6 #include<list>
  7 #include<map>
  8 #include<set>
  9 #include<deque>
 10 #include<stack>
 11 #include<bitset>
 12 #include<algorithm>
 13 #include<functional>
 14 #include<numeric>
 15 #include<utility>
 16 #include<sstream>
 17 #include<iostream>
 18 #include<iomanip>
 19 #include<cstdio>
 20 #include<cmath>
 21 #include<cstdlib>
 22 #include<cstring>
 23 #include<ctime>
 24 #define LL long long
 25 #define maxn 100005
 26 using namespace std;
 27 LL a[maxn];
 28 LL b[maxn];
 29 int la[maxn];
 30 int dp[maxn];
 31 map <LL,int> mp;
 32 struct node{
 33    int l ,r; 
 34    int mx;
 35    int site;; 
 36 }tree[maxn << 2];
 37 int L(int x)
 38 {
 39   return 2 * x;
 40 }
 41 int R(int x)
 42 {
 43   return  2 * x + 1 ;
 44 }
 45 void build(int c, int l , int r)
 46 {
 47     tree[c].l =  l ; 
 48     tree[c].r = r; 
 49     tree[c].mx = 0 ; 
 50     tree[c].site = 0 ; 
 51     if(l == r)
 52     {
 53        return ; 
 54     }
 55     build(L(c),l,(l+r)/2);
 56     build(R(c),(l+r)/2+1,r);
 57 }
 58 void pushup(int c)
 59 {
 60     if(tree[L(c)].mx > tree[R(c)].mx)
 61     {
 62         tree[c].mx = tree[L(c)].mx; 
 63         tree[c].site = tree[L(c)].site;
 64     }else {
 65         tree[c].mx = tree[R(c)].mx; 
 66         tree[c].site = tree[R(c)].site;
 67     }
 68 }
 69 void update(int c, int k,int st,int v )
 70 {
 71     if(tree[c].l == tree[c].r  && tree[c].l == k )
 72     {
 73         tree[c].mx = v ; 
 74         tree[c].site = st ; 
 75         return ;
 76     }
 77     if(k <= (tree[c].l + tree[c].r)/2)
 78     {
 79         update(L(c),k,st,v);
 80     }else {
 81         update(R(c),k,st,v);
 82     }
 83     pushup(c);
 84 }
 85 int mx ;
 86 int site; 
 87 void query(int c, int l ,int r)
 88 {
 89       if(tree[c].l>= l && tree[c].r <= r)
 90       {
 91          if(tree[c].mx > mx )
 92          {
 93             mx = tree[c].mx;
 94             site = tree[c].site ; 
 95          }
 96          return;
 97       }
 98       int m = (tree[c].l + tree[c].r) /2;
 99       if(l <= m)
100          query(L(c),l,r);
101       if(r > m)
102          query(R(c),l,r);
103 }
104 int n , d; 
105 int find(LL x)
106 {
107     int l = 1; 
108     int r = n; 
109     while(l <= r )
110     {
111        int m = (l + r)/2;
112        if(b[m] <= x)
113        {
114           l = m + 1; 
115        }else {
116           r = m - 1;
117        }
118     }
119     return l;
120 }
121 void solve(int k )
122 {
123    if(k == 0 )
124        return;
125    solve(la[k]);
126    printf("%d ",k);
127 }
128 int main(){
129     scanf("%d %d",&n,&d);
130     for(int i =1 ;i <= n;i ++)
131     {
132         scanf("%I64d",&a[i]);
133         b[i] = a[i];
134     }
135     sort(b+1,b+1+n);
136     mp.clear();
137     b[0] = 0 ; 
138     int num = 0 ; ;
139     for(int i = 1;i <= n;i ++)
140     {
141         if(b[i] != b[i-1])
142         {
143             num ++ ;
144              mp[b[i]] = num;
145         }
146     }
147     build(1,1,num);
148     //printf("****%d
",num);
149     int p = 0 ;
150     int pmx =0; 
151     for(int i =1;i <= n;i ++)
152     {
153        int t = mp[a[i]];
154        int l = find(a[i] - d);
155        int r = find(a[i] + d - 1);
156        
157        mx = 0; 
158        site = 0 ;
159        if(l != 1)
160        {
161          //printf("mp[b[l-1]] = %d
",mp[b[l-1]]);
162          query(1,1,mp[b[l-1]]);
163        }
164        if(r != n + 1)
165        {
166           //printf("mp[b[r]] = %d %d
",mp[b[r]],r);
167           query(1,mp[b[r]],num);
168        }
169        la[i] = site;
170        //if(mx != 0 || i == 1)
171           mx ++ ; 
172        
173        dp[i] = mx;
174        if(mx > pmx)
175        {
176          pmx = mx ; 
177          p = i ;    
178        }
179        //printf("%d %d %d
",t,i,mx);
180        update(1,t,i,mx); 
181       // printf("%d %d %d %d
",t,mx,site,tree[1].mx);
182     }
183     
184     printf("%d
",pmx);
185     solve(p);
186     /*printf("
");
187     for(int i = 1;i <= n;i ++)
188         printf("%d ",la[i]);
189     
190         */
191     return 0;
192 }

View Code

F：用线段树统计区间 gcd ，新建一个结构体储存值和下标，找到gcd 以后用两次二分找到区间内 gcd 值的个数即可

  1 // File Name: f.cpp
  2 // Author: darkdream
  3 // Created Time: 2014年10月07日 星期二 21时25分35秒
  4 
  5 #include<vector>
  6 #include<list>
  7 #include<map>
  8 #include<set>
  9 #include<deque>
 10 #include<stack>
 11 #include<bitset>
 12 #include<algorithm>
 13 #include<functional>
 14 #include<numeric>
 15 #include<utility>
 16 #include<sstream>
 17 #include<iostream>
 18 #include<iomanip>
 19 #include<cstdio>
 20 #include<cmath>
 21 #include<cstdlib>
 22 #include<cstring>
 23 #include<ctime>
 24 #define LL long long
 25 #define maxn 100005
 26 using namespace std;
 27 struct node{
 28    int l , r ; 
 29    int gcd;
 30 }tree[maxn << 2];
 31 int a[maxn];
 32 struct node1{
 33     int v,p; 
 34 }b[maxn];
 35 int gcd(int x, int y )
 36 {
 37    return y == 0 ? x:gcd(y,x%y);
 38 }
 39 int L(int x)
 40 {
 41    return 2 * x ; 
 42 }
 43 int R(int x)
 44 {
 45    return 2 * x + 1 ; 
 46 }
 47 void pushup(int c)
 48 {
 49    tree[c].gcd = gcd(tree[L(c)].gcd,tree[R(c)].gcd);
 50 }
 51 void build(int c, int l , int r)
 52 {
 53    tree[c].l = l ;
 54    tree[c].r = r;
 55    //printf("%d %d %d
",c,l,r);
 56    if(l == r)
 57    {
 58       tree[c].gcd = a[l];
 59       return ;
 60    }
 61    int m = (l + r)/2;
 62    build(L(c),l,m);
 63    build(R(c),m+1,r);
 64    pushup(c);
 65 }
 66 int cmp(node1 a, node1 b)
 67 {
 68   if(a.v == b.v)
 69   {
 70     return a.p < b.p ;
 71   }
 72   return a.v < b.v ; 
 73 }
 74 int ans = 0 ;
 75 
 76 void query(int c, int l, int r)
 77 {
 78     //printf("%d %d %d %d %d 
",c,l,r,tree[c].l ,tree[c].r);
 79     if(tree[c].l >= l && tree[c].r <= r)
 80     {
 81          ans = gcd(ans,tree[c].gcd);
 82          return ;
 83     }
 84     int m = (tree[c].l + tree[c].r)/2;
 85     if(l <= m)
 86          query(L(c),l,r);
 87     if(r > m )
 88          query(R(c),l,r);
 89 }
 90 int n ;
 91 int  find(int x)
 92 {
 93     int l = 1; 
 94     int r = n ; 
 95     while(l <= r)
 96     {
 97       int m = (l +r )/2;
 98       if(b[m].v <= x)
 99       {
100          l = m + 1;  ; 
101       }else {
102          r = m - 1; 
103       }
104     }
105     return l ; 
106 }
107 int ll, rr ;
108 int find1(int x)
109 {
110     int l = ll ;
111     int r = rr ; 
112     while(l <= r)
113     {
114       int m = (l +r )/2;
115       if(b[m].p <= x)
116       {
117          l = m + 1;  ; 
118       }else {
119          r = m - 1; 
120       }
121      }
122    return l ;
123 }
124 int main(){
125    scanf("%d",&n);
126    for(int i =1;i <= n;i ++)
127    {
128       scanf("%d",&a[i]);
129       b[i].v = a[i];
130       b[i].p = i;
131    }
132    build(1,1,n);
133    sort(b+1,b+n+1,cmp);
134    //for(int i =1 ;i <= n;i ++)
135      //  printf("%d %d
",b[i].v,b[i].p);
136    int m ; 
137    scanf("%d",&m);
138    for(int i = 1;i <= m;i ++)
139    {
140      int l , r ; 
141      scanf("%d %d",&l,&r);
142      ans = a[l];
143      query(1,l,r);
144      //printf("%d
",ans);
145      ll = find(ans-1);
146      if(b[ll].v != ans)
147      {
148          printf("%d
",r-l+1);
149          continue;
150      }else{
151          rr = find(ans);
152          rr --;
153          int x, y ; 
154          x = find1(l-1);
155          if(b[x].p > r)
156          {
157             printf("%d
",r-l+1);
158             continue;
159          }else{
160             y = find1(r);
161             y -- ;
162             printf("%d
",r-l+1 - (y-x+1));
163          } 
164      }
165 
166    }
167 return 0;
168 }

View Code

没有梦想，何谈远方

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原文地址：https://www.cnblogs.com/zyue/p/4009555.html