Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible from the N (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weight Wi (1 ≤ Wi ≤ 400), a 'desirability' factor Di (1 ≤ Di ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M (1 ≤ M ≤ 12,880).
Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.
* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: Wi and Di
* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints
4 6 1 4 2 6 3 12 2 7Sample Output
23
题意：给n种物品的体积和价值，在V体积下所能装的最大价值是多少，01背包简单问题。
#include<stdio.h>
#include<string.h>
int Max(int x,int y)
{if(x>y)return x;return y;
}
int main()
{int i,j,k,m,n,v;int f[5000],w[5000];int c[15000]; while(scanf("%d%d",&m,&v)!=EOF){memset(c,0,sizeof(c));memset(f,0,sizeof(f));memset(w,0,sizeof(w)); for(i=1;i<=m;i++)scanf("%d%d",&f[i],&w[i]);for(i=1;i<=m;i++){for(j=v;j>=f[i];j--){c[j]=Max(c[j],c[j-f[i]]+w[i]);//动态方程，对于每个体积状态下判断当前体积上能否到最大价值}}printf("%d\n",c[v]);}return 0;
}