Food Problem Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1243    Accepted Submission(s): 368 Problem Description Few days before a game of orienteering, Bell came to a mathematician to sol…

http://acm.hdu.edu.cn/showproblem.php?pid=2191 New~ 欢迎“热爱编程”的高考少年——报考杭州电子科技大学计算机学院关于2015年杭电ACM暑期集训队的选拔 悼念512汶川大地震遇难同胞——珍惜现在,感恩生活 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 17930    Accepted…

Big Event in HDU Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 32578    Accepted Submission(s): 11377 Problem Description Nowadays, we all know that Computer College is the biggest department…

Big Event in HDU Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 19108    Accepted Submission(s): 6707 Problem Description Nowadays, we all know that Computer College is the biggest department…

Big Event in HDU   Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1139 Accepted Submission(s): 444 Problem Description Nowadays, we all know that Computer College is the biggest department in HD…

http://acm.hdu.edu.cn/showproblem.php?pid=1171 Big Event in HDU Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 28483    Accepted Submission(s): 10027 Problem Description Nowadays, we all know…

这里;http://acm.hdu.edu.cn/showproblem.php?pid=1059 题意是有价值分别为1,2,3,4,5,6的商品各若干个,给出每种商品的数量,问是否能够分成价值相等的两份. 联想到多重背包,稍微用二进制优化一下.(最近身体不适,压力山大啊) #include<iostream> #include<cstring> #include<cstdio> #define inf 70000 using namespace std; int dp…

Problem Description Nowadays, we all know that Computer College is the biggest department in HDU. But, maybe you don't know that Computer College had ever been split into Computer College and Software College in 2002.The splitting is absolutely a big…

HDU 1114 Piggy-Bank 完全背包问题. 想想我们01背包是逆序遍历是为了保证什么? 保证每件物品只有两种状态,取或者不取.那么正序遍历呢? 这不就正好满足完全背包的条件了吗 means:给出小猪钱罐的重量和装满钱后的重量,然后是几组数据,每组数据包括每种钱币的价值与重量要求出装满钱罐时的最小价值 #include<cstdio> #include<cstring> #include<cmath> #include<iostream> usin…

题意:价值分别为1,2,3,4,5,6的物品个数分别为a[1],a[2],a[3],a[4],a[5],a[6],问能不能分成两堆价值相等的. 解法:转化成多重背包 #include<stdio.h> #include<string.h> #include<algorithm> #include<iostream> using namespace std; ]; ]; int nValue; //0-1背包,代价为cost,获得的价值为weight void…