/ 第一个多重背包题目 真的不理解二进制优化 /http://acm.hdu.edu.cn/webcontest/contest_showproblem.php?cid=10594&pid=1001&ojid 以下是用二进制优化的 不会超时 include include include include using namespace std; int dp[200000],a[7]; void zeroOne(int weiht,int value,int c) { for(int j=…

Dividing Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 20635    Accepted Submission(s): 5813 Problem Description Marsha and Bill own a collection of marbles. They want to split the collection…

多重背包模板- #include <stdio.h> #include <string.h> int a[7]; int f[100005]; int v, k; void ZeroOnePack(int cost, int weight) { for (int i = v; i >= cost; i--) if (f[i - cost] + weight > f[i]) f[i] = f[i - cost] + weight; } void CompletePack(…

点击打开链接链接 Dividing Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 17544    Accepted Submission(s): 4912 Problem Description Marsha and Bill own a collection of marbles. They want to split the c…

Dividing 给出n个物品的价值和数量,问是否能够平分.…

Dividing Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 63980   Accepted: 16591 Description Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbl…

Q: 倍增优化后, 还是有重复的元素, 怎么办 A: 假定重复的元素比较少, 不用考虑 Description Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbles. This would be easy if all the marbles had the…

Problem Description Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbles. This would be easy if all the marbles had the same value, because then they could…

link:http://acm.hdu.edu.cn/showproblem.php?pid=1059 最简单的那种 #include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <cctype> #include <algorithm> #include <queue>…

来源poj1059 Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbles. This would be easy if all the marbles had the same value, because then they could just split…