两组牌中两张牌相比能赢的就连,后求最大匹配. #include <cmath> #include <cstdlib> #include <cstdio> #include <cstring> #include <algorithm> #include <fstream> #include <iostream> #define rep(i, l, r) for(int i=l; i<=r; i++) #define c…

http://acm.hdu.edu.cn/showproblem.php?pid=1528 Card Game Cheater Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1559    Accepted Submission(s): 820 Problem Description Adam and Eve play a card…

Card Game Cheater Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1072    Accepted Submission(s): 564 Problem Description Adam and Eve play a card game using a regular deck of 52 cards. The rule…

称号: Card Game Cheater Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 103 Accepted Submission(s): 74   Problem Description Adam and Eve play a card game using a regular deck of 52 cards. The rules…

版权声明:来自: 码代码的猿猿的AC之路 http://blog.csdn.net/ck_boss https://blog.csdn.net/u012797220/article/details/35236457 简单二分图匹配.... Card Game Cheater Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1073   …

Card Game Cheater Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1822    Accepted Submission(s): 998 Problem Description Adam and Eve play a card game using a regular deck of 52 cards. The rule…

Card Game Cheater Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1272    Accepted Submission(s): 675 Problem Description Adam and Eve play a card game using a regular deck of 52 cards. The rule…

Card Game Cheater Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1566    Accepted Submission(s): 822 Problem Description Adam and Eve play a card game using a regular deck of 52 cards. The rule…

Card Game Cheater Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1576    Accepted Submission(s): 830 Problem Description Adam and Eve play a card game using a regular deck of 52 cards. The rule…

题目:点击打开链接 题意:两个人纸牌游戏,牌大的人得分.牌大:2 < 3 < 4 < 5 < 6 < 7 < 8 < 9 < T < J < Q < K < A .值一样看花色, hearts (红心) > spades (黑桃) > diamond (方块) > clubs (梅花).问Eve 能得多少分.(每次得1分) 分析:将Eve的每张牌与Adam的所有牌比较,与所有比这张牌小的Adam的牌连边. 然后求最大…

水题,感觉和田忌赛马差不多 #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std; struct P1 { int Num; int Hua; } Play1[]; struct P2 { int Num; int Hua; } Play2[]; bool cmp1(const P1&a,const P1&b) {…

http://acm.split.hdu.edu.cn/showproblem.php?pid=4336 Card Collector Special Judge Problem Description   In your childhood, do you crazy for collecting the beautiful cards in the snacks? They said that, for example, if you collect all the 108 people i…

http://acm.hdu.edu.cn/showproblem.php?pid=4336 题意:n张卡片,每一次取一个盒子,盒子里装有卡片i的概率是p[i],求得到所有卡片所需要开的盒子的期望数(n<=20) #include <cstdio> #include <cstring> using namespace std; const int N=22; int n; double p[N], f[1<<N]; int main() { while(~scan…

这题是一个字符串模拟水题,给12级学弟学妹们找找自信的,嘿嘿; 题目意思就是要你讲身份证的上的省份和生日解析出来输出就可以了: http://acm.hdu.edu.cn/showproblem.php?pid=2629 #include <iostream> #include <string> using namespace std; int main() { int n,t; string home,a; cin>>n; ; k < n ; k++) { ci…

In your childhood, do you crazy for collecting the beautiful cards in the snacks? They said that, for example, if you collect all the 108 people in the famous novel Water Margin, you will win an amazing award. As a smart boy, you notice that to win t…

思路: 分析:假设取的牌顺序是一个序列,那么这种序列在末尾为1时是和取牌序列一一对应的,且是符合“游戏结束时牌恰好被取完”的一种情况. 简证:1.在序列中,任一数 i 的后一个数 j 是必然要放在第 i 堆里的.而值为 i 的数有 a[i]个,所以在 i 后面的数也恰好a[i]个,所以a[i]个数被放到第 i 堆,符合题目约束条件. 2.在序列中,由于游戏是从第一堆开始的,所以第一个数虽然没有前驱,但是他是放在第 1 堆的.所以如果 1 不为最后一个数,那么第一堆中必然有a[1]+1个数了,不行…

二分图最大匹配问题 扑克题还是用map比较方便 #include<bits/stdc++.h> using namespace std; #define MAXI 52 ]; ]; int used[MAXI]; int vis[MAXI]; int n,m; map<char,int>mm; bool judge(int a,int b) { ]]>mm[ yd[b][] ] ) return true; ]]==mm[ yd[b][] ]) ]]>mm[ yd[b]…

题意与分析 题意是这样的:有\(n\)张牌,然后第一行是Adam的牌,第二行是Eve的牌:每两个字符代表一张牌,第一个字符表示牌的点数,第二个表示牌的花色.Adam和Eve每次从自己的牌中选出一张牌进行比较,先看大小再看花色,花色顺序是C,D,S,H(依次增大),谁的牌大谁就加一分,问Eve最多能得到多少分. 思路:最大二分匹配,Eve的牌为集合1,Adam的牌为集合2,集合1中的牌与集合2中比它小的牌建立联系,找最大匹配.这种两个集合的要想到试试二分图. 代码 /* ACM Code writ…

#include<stdio.h> #include<string.h> int map[100][100],mark[100],link[100],max2,k; int find(int u) {  int i;  for(i=1;i<=k;i++) {   if(!mark[i]&&map[u][i]) {    mark[i]=1;    if(!link[i]||find(link[i])) {     link[i]=u;     return 1…

简单题 给出身份证号 判断住址 和出生年月 熟练字符串的操作 主要是string::substr(s, l)//s:起始位置 l长度 #include <iostream> #include <stdio.h> #include <string> #include <string.h> #include <map> #include <fstream> using namespace std; map<string,string…

转自——http://blog.csdn.net/qwe20060514/article/details/8112550 =============================以下是最小生成树+并查集======================================[HDU]1213   How Many Tables   基础并查集★1272   小希的迷宫   基础并查集★1325&&poj1308  Is It A Tree?   基础并查集★1856   More i…

转载:from http://blog.csdn.net/qq_28236309/article/details/47818349 基础题:1000.1001.1004.1005.1008.1012.1013.1014.1017.1019.1021.1028.1029. 1032.1037.1040.1048.1056.1058.1061.1070.1076.1089.1090.1091.1092.1093. 1094.1095.1096.1097.1098.1106.1108.1157.116…

=============================以下是最小生成树+并查集====================================== [HDU] 1213 How Many Tables 基础并查集★ 1272 小希的迷宫 基础并查集★ 1325&&poj1308 Is It A Tree? 基础并查集★ 1856 More is better 基础并查集★ 1102 Constructing Roads 基础最小生成树★ 1232 畅通工程 基础并查集★ 123…

=============================以下是最小生成树+并查集====================================== [HDU] 1213 How Many Tables 基础并查集★ 1272 小希的迷宫 基础并查集★ 1325&&poj1308 Is It A Tree? 基础并查集★ 1856 More is better 基础并查集★ 1102 Constructing Roads 基础最小生成树★ 1232 畅通工程 基础并查集★ 123…

原地址:http://www.byywee.com/page/M0/S607/607452.html 总结了一下ACM STEPS的各章内容,趁便附上我的Steps题号(每人的不一样). 别的,此文首要目标是为了装逼: 大牛请疏忽: 摸索欲斗劲强的请疏忽: 其实不乐于从A+B刷起的可以找到须要的响应题号操练-- 1.1 根蒂根基输入输出:LCY的 A+B 8题 (1089~1096) 1.2 C说话根蒂根基:根蒂根基入门题 (2104,2088,1076,2095,1061,1170,3361,…

各种杂题,水题,模拟,包括简单数论. 1001 A+B 1002 A+B+C 1009 Fat Cat 1010 The Angle 1011 Unix ls 1012 Decoding Task 1019 Grandpa's Other Estate 1034 Simple Arithmetics 1036 Complete the sequence! 1043 Maya Calendar 1054 Game Prediction 1057 Mileage Bank 1067 Rails 10…

杭电ACM分类: 1001 整数求和 水题1002 C语言实验题——两个数比较 水题1003 1.2.3.4.5... 简单题1004 渊子赛马 排序+贪心的方法归并1005 Hero In Maze 广度搜索1006 Redraiment猜想 数论:容斥定理1007 童年生活二三事 递推题1008 University 简单hash1009 目标柏林 简单模拟题1010 Rails 模拟题(堆栈)1011 Box of Bricks 简单题1012 IMMEDIATE DECODABILITY…

D - 粉碎叛乱 Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1528 Description Adam and Eve play a card game using a regular deck of 52 cards. The rules are simple. The players sit on opposite sides…

=============================以下是最小生成树+并查集====================================== [HDU] How Many Tables 基础并查集★ 小希的迷宫 基础并查集★ &&poj1308 Is It A Tree? 基础并查集★ More is better 基础并查集★ Constructing Roads 基础最小生成树★ 畅通工程 基础并查集★ 还是畅通工程 基础最小生成树★ 畅通工程 基础最小生成树★ 畅通…

数学期望 P=Σ每一种状态*对应的概率. 因为不可能枚举完所有的状态,有时也不可能枚举完,比如抛硬币,有可能一直是正面,etc.在没有接触数学期望时看到数学期望的题可能会觉得很阔怕(因为我高中就是这么认为的,对不起何老板了QwQ),避之不及. 但是现在发现大多数题就是手动找公式或者DP推出即可,只要处理好边界,然后写好方程,代码超级简短.与常规的求解不同,数学期望经常逆向推出. 比如常规的dp[x]可能表示到了x这一状态有多少,最后答案是dp[n].而数学期望的dp[x]一般表示到了x这一状态还…