题目描述：
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.
有N件物品和一个容量为V的背包。第i件物品的重量是c[i]，价值是w[i]。求解将哪些物品装入背包可使这些物品的重量总和不超过背包容量，且价值总和最大。
输入格式：
* 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
第一行：物品个数N和背包大小M
第二行至第N+1行：第i个物品的重量C[i]和价值W[i]
输出格式：
* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints
输出一行最大价值。
输入输出样例：
输入 #1复制
4 6 1 4 2 6 3 12 2 7输出 #1复制
23
AC Code：
#include<bits/stdc++.h>
using namespace std;
#define N 15000
int m,n,dp[N],weight[N],value[N];
int main() {
scanf("%d %d",&n,&m);
for(int i=1;i<=n;i++) {
scanf("%d %d",&weight[i],&value[i]);
}
for(int i=1;i<=n;i++) {
for(int j=m;j>=weight[i];j--) {
dp[j]=max(dp[j],dp[j-weight[i]]+value[i]);
}
}
printf("%d\n",dp[m]);
return 0;
}