#include<cstdio> #include<cstring> using namespace std; +; ]; int dp[SIZE]; bool check() { ;i<;i++) ) return true; return false; } int sum; int main() { ; while(true) { sum=; ;i<;i++) { scanf("%d",&a[i]); sum+=(i+)*a[i]; }…

描述 http://poj.org/problem?id=1742 n种不同面额的硬币 ai ,每种各 mi 个,判断可以从这些数字值中选出若干使它们组成的面额恰好为 k 的 k 的个数. 原型: n种不同大小的数字 ai ,每种各 mi 个,判断是否可以从这些数字之中选出若干使它们的和恰好为 k . Coins Time Limit: 3000MS   Memory Limit: 30000K Total Submissions: 33732   Accepted: 11453 Descrip…

4013多重部分和问题 难度级别:B: 运行时间限制:2000ms: 运行空间限制:262144KB: 代码长度限制:2000000B 试题描述 n种大小不同的数字 Ai,每种各Mi个,判断是否可以从这些数字之中选出若干个使他们的和恰好为K. 输入 第一行为两个正整数n,K.第二行为n个数Ai,以空格隔开.第三行为n个数Mi,以空格隔开. 输出 若可行则输出"yes"否则输出"no" 输入示例 3 173 5 83 2 2 输出示例 yes 其他说明 1<=n…

1.解法一:多重背包 #include<iostream> #include<cstdio> using namespace std; #define MAX(a,b) (a>b)?a:b +; int dp[SIZE]; ]; int nLimit; void ZeroOnePack(int cost, int value) { for(int i=nLimit; i>=cost; i--) dp[i]=MAX(dp[i],dp[i-cost]+value); } v…

多重部分和问题 代码(C) 本文地址: http://blog.csdn.net/caroline_wendy 题目: 有n种不同大小的数字a, 每种各m个. 推断能否够从这些数字之中选出若干使它们的和恰好为K. 使用 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 起始 0 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 0(3,3) 3 -1 -1 2 -1 -1 1 -1 -1 0 -1 -1 -1…

Coins Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10961    Accepted Submission(s): 4418 Problem Description Whuacmers use coins.They have coins of value A1,A2,A3...An Silverland dollar. One…

Description People in Silverland use coins.They have coins of value A1,A2,A3...An Silverland dollar.One day Tony opened his money-box and found there were some coins.He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact pri…

Problem Description Whuacmers use coins.They have coins of value A1,A2,A3...An Silverland dollar. One day Hibix opened purse and found there were some coins. He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact price(witho…

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…

当初学者最开始学习 dp 的时候往往接触的是一大堆的 背包 dp 问题, 那么我们在这里就不妨讨论一下常见的几种背包的 dp 问题: 初级的时候背包 dp 就完全相当于BFS DFS 进行搜索之后的记忆化查找. 背包问题 一 .   0 ~ 1 背包问题 实现一. return max ( rec ( i + 1, j ) , rec ( i + 1, j - cost[ i ]) + value[ i ]); //对第 i  件物品的选或者不选 记忆化.      这个是由于实现一产生的dp数…