链接：https://www.nowcoder.com/acm/contest/96/A
来源：牛客网
“LL是什么？这都不知道的话，别说自己是程序猿啊！”
“longlong？”
“。。。肯定是LoveLive啊！”
qwb为了检验你是否是真正的程序猿，决定出道题考考你：现在程序会输入一行字符串，如果恰好是lovelive（不区分大小写）就输出yes，否则输出no。
“longlong？”
“。。。肯定是LoveLive啊！”
qwb为了检验你是否是真正的程序猿，决定出道题考考你：现在程序会输入一行字符串，如果恰好是lovelive（不区分大小写）就输出yes，否则输出no。
输入描述:
输入有多组（组数不超过100），每组输入一行字符串（字符串长度不超过100）。
输出描述:
输出占一行，如果输入符合要求则输出yes，否则输出no。
示例1
输入
longlong LoveLive love live
输出
no yes no
分析：主要是想了解一下C++中tolower语句
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链接：https://www.nowcoder.com/acm/contest/96/C
来源：牛客网
题目描述
durong有a台iphonex和b台s8，并且放在一个保险箱里，durong现在一台一台从保险箱随机拿出这些手机，现在他想知道第k次拿出s8的概率是多少
输入描述:
第一行一个正整数T，表示数据组数。(1<=T<=10000)
接下来T行输入a，b，k
其中（1<=a,b,k<=1e9) k<=a+b;
输出描述:
第k次拿出s8的概率，保留三位小数
示例1
输入
1 1 1 1
输出
0.500
分析：前面拿球不影响后面，所以第k次拿球是C(b, 1)，总排列组合是A(a + b, a + b),拿去之后是A(a + b -1 , a + b - 1);
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define FF(i, a, b) for(int i = a; i < b; i++) 5 #define RR(i, a, b) for(int i = a; i > b; i++) 6 #define ME(a, b) memset(a, b, sizeof(a)) 7 #define SC(x) scanf("%d", &x) 8 #define PR(x) printf("%d ", x) 9 #define MAX 1000 10 #define M 8 11 #define N 6 12 typedef long long LL; 13 typedef pair<int, int> Author; 14 15 int main(void){ 16 int T, a, b, k; 17 18 cin>>T; 19 while(T--){ 20 cin>>a>>b>>k; 21 cout<<fixed<<setprecision(3)<<b * 1.0 / (a + b) * 1.0<<endl; 22 } 23 24 return 0; 25 }
链接：https://www.nowcoder.com/acm/contest/96/D
来源：牛客网
时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld
题目描述
离散数学中有种名叫“哈斯图”的东西。
在这题中，你们需要计算的是一些正整数在偏序关系“整除”下的哈斯图的边数。用大白话讲，在偏序关系“整除”下的哈斯图，就是把一个个正整数看成一个个图的节点，某些节点之间有边。连边的规则是这样的：对于任意两个正整数a和b(a<b)来说，如果b%a==0,并且不存在一个正整数c（a<c<b），使得条件b%c==0和c%a==0同时成立，那么我们就在节点a和节点b之间连一条边。
现在问题是，给你们2个数L，R(1<=L,R<=1e6)。求由L，L+1，L+2...R这R-L+1个正整数在偏序关系“整除”下的哈斯图的边数。
比如L=1，R=4,节点的组合有(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)。组合（1,2），（1,3），（2,4）可以连边。（1,4）因为中间存在c=2，不符合连边条件。所以当L=1，R=4的时候，这个哈斯图有3条边。
输入描述:
多组输入，不超过1000组数据
每组数据一行，包含2个正整数L和R，中间由空格分开。
输出描述:
每组数据输出一行，包含一个整数表示哈斯图的边数。
示例1
输入
1 4 4 10 1 10
输出
3 2 11
分析：就是a,c中间不能有b能被C, a整除，筛选素数，判断p[0]是为了控制边界，再计算p[i]中的素数即可
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链接：https://www.nowcoder.com/acm/contest/96/E
来源：牛客网
时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld
题目描述
AA的欧尼酱qwb是个考古学家，有一天qwb发现了只白白圆圆小小的木乃伊，它是个爱哭鬼却很努力。qwb想把这么可爱的小木乃伊送给
AA，于是便找上了快递姐姐，这下可让快递姐姐犯愁了，因为去往AA家的路实在太难走了（甚至有可能没有路能走到AA家），快递姐姐
找上聪明的ACMer，想请你帮忙找出最快到达AA家的路，你行吗？
AA，于是便找上了快递姐姐，这下可让快递姐姐犯愁了，因为去往AA家的路实在太难走了（甚至有可能没有路能走到AA家），快递姐姐
找上聪明的ACMer，想请你帮忙找出最快到达AA家的路，你行吗？
输入描述:
第一行输入两个整数n和m（2<=n<=m<=200000），分别表示有n座城市和m条路，城市编号为1~n（快递姐姐所在城市为1，AA所在城市为n）。
接下来m行，每行输入3个整数u,v,w（u,v<=n，w<=100000），分别表示城市u和城市v之间有一条长为w的路。
输出描述:
输出结果占一行，输出快递姐姐到达AA家最短需要走多远的路，如果没有路能走到AA家，则输出“qwb baka”（不用输出双引号）。
示例1
输入
4 4 1 2 1 2 3 2 3 4 3 2 3 1
输出
5
分析：看到神仙的DP解题，也有神仙用Spfa和Dijkstra过。
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LBBgGgZHsr;vrr;rcXpP5UUS5X5S2XS5SS1URBggpEZgDDGOGggRD 57 ,;irrs2KgQRPJJJSUXKXSHJpBM . . . ZBBOUsr;:. ,,,rHgZPUaSaUSUSSXaUPXsSBB6ZZp6EEGOOGDEggO 58 ;BBQQZaSJ5J15SJDQ7 .iEBL:. . .,:,:;SRQGZXPXXSHaS2UUSUSswQB6OZgG6pZZGODEOGDE 59 JgUri1aGEpEpXSS5LBQ, .rLap2Jr ;712OO6ZggRDZXHXpKK2211LLLc7wgBa2PODRGE6OZgDgGDGD 60 7QB; ,Jc76DaZXOZgDPEBBX, :c1HEZ1c;:,.,:;cHDgQQggMZgGOZO6OEGpXsLLsJXHPOBBBc;vss5EMggZDODOgED6 61 LZJ::iwrrr72EPgXU5OBBBBBBBp7: .i5P6wL;, .:LS6DMgMgZKgEOPOGGDDGEXULvLSOBBBBBQBgDQDJ5rrr7JOQQDDZGEgDD 62 wp rR5, .HQRX7r72RQBBBBBQBMEK6Uc7sc7cHOU:, .:LHgOQg6ZOPZpZGGEDERZPwcr7LKDDaJr:. rBX;;:r:ir1DBQMgRgDp 63 7D,sw: , Z. .LgBBQBBOsJaQBBBBBBXrr. .:rwZBBQgROgDgDOERRMgROOScrLsPP5r, 7Qs,::::::rUgRgZOG6 64 ,Q57:r ,: ,r U: .;; .vL7L;r:UG7.:sr;JSgQBgDEZXXUSXpHa21svr7r7s2v:. iQK...:,:::;LLJJws 65 OBHs;. :;;,v H: :6X; rpL;7pSrrr7r;::,....,,:;iirrvvvr, :BBPs;;:,,,,::;:: 66 ::rPDEL:..7:c;7r pK .: :KL.:r:.. ..:icsS5sr:,. JgGgBQDOSJss77r 67 :r;::: , PH r, ,;5: ,::::;7Ls7Lc7;: ,:7JP17rJUs 68 .:J: :; .,:K6BQS7:.,.,. 69 :r7Ji;r7;::;;. 70 . **/ 71 72 #include <bits/stdc++.h> 73 using namespace std; 74 75 #define FF(i, a, b) for(int i = a; i < b; i++) 76 #define RR(i, a, b) for(int i = a; i > b; i++) 77 #define ME(a, b) memset(a, b, sizeof(a)) 78 #define SC(x) scanf("%d", &x) 79 #define PR(x) printf("%d ", x) 80 #define INF 0x3f3f3f3f 81 #define MAX 200005 82 #define MAXNUM 1e11 83 #define MOD 1e9+7 84 #define E 2.71828182845 85 #define PI 3.1415926 86 #define M 8 87 #define N 6 88 typedef long long LL; 89 typedef pair<int, int> Author; 90 vector<pair<string, int> > VP; 91 92 struct Node{ 93 LL u, v, w; 94 }; 95 LL DP[MAX]; 96 97 int main(void){ 98 ios::sync_with_stdio(false); cin.tie(0); 99 int i, n, m, flag, uu, vv, ww; 100 vector<Node> v;Node node; 101 ME(DP, 125); DP[1] = 0;flag = 1; 102 103 cin>>n>>m; 104 for(i = 0; i < m; i++){ 105 cin>>node.u>>node.v>>node.w; 106 v.push_back(node); 107 } 108 109 while(flag){ //²»¶Ï¼ì²é£¬ÕÒµ½×îÓÅ·¾¶ 110 flag = 0; 111 for(i = 0; i < v.size(); i++){ 112 uu = v[i].u;vv = v[i].v; ww = v[i].w; 113 //Ë«Ïò¼ì²é 114 if(DP[uu] + ww < DP[vv]){ 115 DP[vv] = ww + DP[uu]; 116 flag = 1; 117 } 118 if(DP[vv] + ww < DP[uu]){ 119 DP[uu] = ww + DP[vv]; 120 flag = 1; 121 } 122 } 123 } 124 if(DP[n] > MAXNUM) cout<<"qwb baka"<<endl; 125 else cout<<DP[n]<<endl; 126 127 return EXIT_SUCCESS; 128 }
链接：https://www.nowcoder.com/acm/contest/96/G
来源：牛客网
时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld
题目描述
给你一个n*m的图，地图上'.'代表可以走的地方，而'#'代表陷阱不能走，
'P'代表人物位置，'K'代表钥匙，'E'代表出口。人物一个，钥匙有多个，
（'K'的数量<=50)）,出口一个，每个位置可以向（上，下，左，右）四个
方向走一格，花费一个单位时间，现在你需要花费最少的时间拿到钥匙
然后从迷宫的出口出去(若没有钥匙，则不能进入迷宫出口所在的格子)。
'P'代表人物位置，'K'代表钥匙，'E'代表出口。人物一个，钥匙有多个，
（'K'的数量<=50)）,出口一个，每个位置可以向（上，下，左，右）四个
方向走一格，花费一个单位时间，现在你需要花费最少的时间拿到钥匙
然后从迷宫的出口出去(若没有钥匙，则不能进入迷宫出口所在的格子)。
输入描述:
第一行一个整数T(T <= 50)，代表数据的组数
接下来一行n,m(n<=500,m<=500),代表地图的行和列
接下来n行，每行一个长度为m的字符串，组成一个图。
输出描述:
如果可以出去，输出所花费的最少时间。
如果不能出去，输出一行"No solution"。
示例1
输入
3 5 5 ....P ##..E K#... ##... ..... 5 5 P.... ..... ..E.. ..... ....K 5 5 P#..E .#.#. .#.#. .#.#. ...#K
输出
No solution 12 No solution
分析：单向BFS+三维数组特判就可以了，也有人用双向BFS找最短路
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JgGgBQDOSJss77r 67 :r;::: , PH r, ,;5: ,::::;7Ls7Lc7;: ,:7JP17rJUs 68 .:J: :; .,:K6BQS7:.,.,. 69 :r7Ji;r7;::;;. 70 . **/ 71 72 #include <bits/stdc++.h> 73 using namespace std; 74 #define FF(i, a, b) for(int i = a; i < b; i++) 75 #define RR(i, a, b) for(int i = a; i > b; i++) 76 #define ME(a, b) memset(a, b, sizeof(a)) 77 #define SC(x) scanf("%d", &x) 78 #define PR(x) printf("%d ", x) 79 #define INF 0x3f3f3f3f 80 #define MAX 1001 81 #define MOD 1e9+7 82 #define E 2.71828182845 83 #define PI 3.1415926 84 #define M 8 85 #define N 6 86 typedef long long LL; 87 typedef pair<int, int> Author; 88 vector<pair<string, int> > VP; 89 90 int n, k; 91 int dis[MAX][MAX][2]; //ÓÐûÓÐÔ¿³× 92 int dir[4][2] = {1, 0, 0, 1, -1, 0, 0, -1}; 93 char matrix[MAX][MAX]; 94 typedef struct Node{ 95 int x, y, flag, step; //step¼ÆËã·³Ì,flagÊÇÓÐÎÞÔ¿³× 96 }node; 97 98 bool check(int x, int y){ 99 if(x >= 0 && x < n && y >= 0 && y < k) return 1; 100 return 0; 101 } 102 103 int BFS(int x, int y){ 104 queue<Node> p;Node temp, temp2;int i; 105 temp.x = x; temp.y = y; temp.step = 0; temp.flag = 0; 106 dis[x][y][0] = 1; //³õʼλÖà 107 108 p.push(temp); 109 while(!p.empty()){ 110 temp2 = p.front(); p.pop(); 111 if(matrix[temp2.x][temp2.y] == 'E') return temp2.step; //Óöµ½ÖÕµã¾ÍÍ˳ö 112 for(i = 0; i < 4; i++){ 113 int xx = temp2.x + dir[i][0]; 114 int yy = temp2.y + dir[i][1]; 115 116 if(check(xx, yy) && matrix[xx][yy] != '#'){ 117 Node temp3; 118 if(matrix[xx][yy] == 'K') temp3.flag = 1; //ÕÒµ½Ô¿³× 119 else temp3.flag = temp2.flag; 120 if(dis[xx][yy][temp3.flag]) continue; //Èô¸Ãµã·ÃÎʹý£¬ÔòÖ±½Ócontinue 121 dis[xx][yy][temp3.flag] = 1; 122 if(matrix[xx][yy] =='E' && !dis[xx][yy][1]) continue; //ÈôÕÒµ½ÖÕµãÇÒû±»·ÃÎʹý£¬Ôò½áÊø£¬·ÀÖ¹¾¹ýÖÕµã 123 temp3.x = xx; temp3.y = yy; temp3.step = temp2.step + 1; p.push(temp3); 124 } 125 } 126 } 127 return -1; 128 } 129 130 131 int main(void){ 132 int i, j, T, bx, by; 133 134 cin>>T; 135 while(T--){ 136 ME(dis, 0); 137 cin>>n>>k; 138 for(i = 0; i < n; i++){ 139 cin>>matrix[i]; 140 for(j = 0; j < k; j++)if(matrix[i][j] == 'P'){bx = i; by = j;} 141 } 142 int ret = BFS(bx, by); 143 if(ret != -1) cout<<ret<<endl; 144 else cout<<"No solution"<<endl; 145 } 146 147 return EXIT_SUCCESS; 148 }
链接：https://www.nowcoder.com/acm/contest/96/J
来源：牛客网
时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld
题目描述
一天durong同学买了一个无限长的杯子，同时买了n个球，并且标号为1，2，3......n，durong同学突然想到一个问题----如果他把n个球依次，也就是按照1，2，3...n的顺序放进杯子里，然后在全部拿出来（注意不一定要等到全部放进去才能拿出球），并且会记录放进和拿出球的顺序，
durong想知道，要满足当第m个球进去后，杯子中此时恰好有k个球，然后仍然要把剩下的n-m个球放进去，最后杯中的球要取光，这样的放进和拿出球的顺序有多少种，答案有可能很大，所以mod上1e9+7
durong想知道，要满足当第m个球进去后，杯子中此时恰好有k个球，然后仍然要把剩下的n-m个球放进去，最后杯中的球要取光，这样的放进和拿出球的顺序有多少种，答案有可能很大，所以mod上1e9+7
输入描述:
1<=n,m,k<=1e6（m可能大于n，k可能大于m）
第一行一个正整数T，表示数据组数。(1<=T<=10000)
对于每组数据包含一行三个正整数n,m,k。
输出描述:
对于每组数据输出一个正整数表示答案。
由于答案可能过大，所以只需要输出对1e9+7取模后的答案
示例1
输入
2 3 3 3 3 3 2
输出
1 2
分析：反向思维，第m个球不算，前面共m - 1个球,已放好k个球，需要放m - k 个球,第二步,放n - m个球，取n - m + k个球，卡特兰数公式判断即可，注意数组范围和逆元，不然会段错误
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JgGgBQDOSJss77r 67 :r;::: , PH r, ,;5: ,::::;7Ls7Lc7;: ,:7JP17rJUs 68 .:J: :; .,:K6BQS7:.,.,. 69 :r7Ji;r7;::;;. 70 . **/ 71 72 #include <bits/stdc++.h> 73 using namespace std; 74 75 #define FF(i, a, b) for(int i = a; i < b; i++) 76 #define RR(i, a, b) for(int i = a; i > b; i++) 77 #define ME(a, b) memset(a, b, sizeof(a)) 78 #define SC(x) scanf("%d", &x) 79 #define PR(x) printf("%d ", x) 80 #define INF 0x3f3f3f3f 81 #define MAX 4100000 82 #define MOD 1000000007 83 #define E 2.71828182845 84 #define PI 3.14159265358979 85 #define M 8 86 #define N 6 87 typedef long long LL; 88 typedef pair<int, int> Author; 89 vector<pair<string, int> > VP; 90 91 LL fac[MAX]; 92 //标准快速幂 93 LL qkm(LL a, LL b, LL mod){ 94 LL ans = 1; 95 for(a %= mod; b; b >>= 1, a = a * a % mod)if(b & 1)ans=ans*a%mod; 96 return ans; 97 } 98 99 LL C(LL a, LL b){ 100 return fac[a] * qkm(fac[b], MOD - 2, MOD) % MOD * qkm(fac[a - b], MOD - 2, MOD) % MOD; 101 } 102 103 //卡特兰数 104 LL Catalan(LL n, LL m){ 105 if(m == 0) return 1; 106 else return (C(n + m, n) - C(n + m, n + 1) + MOD) % MOD; //被CF坑的 107 } 108 109 int main(void){ 110 ios::sync_with_stdio(false); cin.tie(0); 111 LL i, T, n, m, k; 112 cin>>T; 113 114 for(fac[0] = 1, fac[1] = 1, i = 2; i <= MAX; i++)fac[i] = (fac[i - 1] * i) % MOD; 115 while(T--){ 116 cin>>n>>m>>k; 117 if(m > n || k > m) cout<<0<<endl; 118 else cout<<(Catalan(m - 1, m - k) * Catalan(n - m + k, n - m))% MOD<<endl; //逆向思维 119 } 120 121 return EXIT_SUCCESS; 122 }
链接：https://www.nowcoder.com/acm/contest/96/L
来源：牛客网
题目描述
集训队暑训的时候，是睡在机房的，有几个小伙子不甘寂寞，带了仓鼠来机房。有的还不止带了一只！
为此，贴心的集训队刘队长买了一大一小两种放仓鼠笼子的架子，大的可以放两个仓鼠笼 小的只能放一个。
要注意的是仓鼠是领地动物，一般来讲 仓鼠笼与仓鼠笼之间是不能贴在一起的，
但同一个人带来的仓鼠的仓鼠笼可以被放在一起（既两个仓鼠笼如果属于同一个人，那么这两个仓鼠笼就可以被放在同一个大架子上）
现在告诉你现有的两种架子的数量和多少人带了仓鼠以及每个人带的仓鼠的数量，问你是否能将这些仓鼠放下。
为此，贴心的集训队刘队长买了一大一小两种放仓鼠笼子的架子，大的可以放两个仓鼠笼 小的只能放一个。
要注意的是仓鼠是领地动物，一般来讲 仓鼠笼与仓鼠笼之间是不能贴在一起的，
但同一个人带来的仓鼠的仓鼠笼可以被放在一起（既两个仓鼠笼如果属于同一个人，那么这两个仓鼠笼就可以被放在同一个大架子上）
现在告诉你现有的两种架子的数量和多少人带了仓鼠以及每个人带的仓鼠的数量，问你是否能将这些仓鼠放下。
输入描述:
第一行为一个数T 表示测试样例组数
对于每组测试数据
第1行输入a,b,n; 分别为小架子的数量，大架子的数量，带了仓鼠的人数
第2行有n个数字P1
-Pn
，分别表示每个人带的仓鼠的数量
输出描述:
如果放得下 则输出"Yes" 否则输出"No"
示例1
输入
1 1 2 4 1 1 1 1
输出
No
分析：数学思维，先对所有人计算最多大架子数，再判断比实际大架子数量多还是少，多就需要双倍小架子，少就需要把剩余大架子数当成小架子计算
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JgGgBQDOSJss77r 67 :r;::: , PH r, ,;5: ,::::;7Ls7Lc7;: ,:7JP17rJUs 68 .:J: :; .,:K6BQS7:.,.,. 69 :r7Ji;r7;::;;. 70 . **/ 71 72 #include <bits/stdc++.h> 73 using namespace std; 74 75 #define FF(i, a, b) for(int i = a; i < b; i++) 76 #define RR(i, a, b) for(int i = a; i > b; i++) 77 #define ME(a, b) memset(a, b, sizeof(a)) 78 #define SC(x) scanf("%d", &x) 79 #define PR(x) printf("%d ", x) 80 #define INF 0x3f3f3f3f 81 #define MAX 1001 82 #define MOD 1e9+7 83 #define E 2.71828182845 84 #define PI 3.1415926 85 #define M 8 86 #define N 6 87 typedef long long LL; 88 typedef pair<int, int> Author; 89 vector<pair<string, int> > VP; 90 91 int main(void){ 92 ios::sync_with_stdio(false); cin.tie(0); 93 LL T, odd, even, i, a, b, n, num, flag; 94 95 cin>>T; 96 while(T--){ 97 cin>>a>>b>>n; odd = even = 0;flag = 0; 98 for(i = 0; i < n; i++){ 99 cin>>num; 100 even += (num / 2); 101 num %= 2; 102 odd += num; 103 } 104 if(b >= even) b -= even; 105 else{ 106 flag = -1; 107 even -= b; 108 } 109 if(flag == 0){ 110 if(a + b >= odd) cout<<"Yes"<<endl; 111 else cout<<"No"<<endl; 112 } 113 else { 114 if(a >= odd + even * 2) cout<<"Yes"<<endl; 115 else cout<<"No"<<endl; 116 } 117 } 118 119 return EXIT_SUCCESS; 120 }