#　题目

## Description

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.

## Input

• 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

## Output

• 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 7

23

# 代码

```/*
By:OhYee
Github:OhYee
HomePage:http://www.oyohyee.com
Email:oyohyee@oyohyee.com

かしこいかわいい？
エリーチカ！

*/

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;

const int maxn = 20000;
int dp[maxn];
int v;

void ZeroOnePack(int cost,int weight) {
for(int i = v; i >= cost; i--)
dp[i] = max(dp[i],dp[i - cost] + weight);
}

bool Do() {
int n,m;
if(scanf("%d%d",&n,&m) == EOF)
return false;

v = m;
memset(dp,0,sizeof(dp));

for(int i = 1;i <= n;i++) {
int a,b;
scanf("%d%d",&a,&b);
ZeroOnePack(a,b);
}

printf("%d\n",dp[m]);
return true;
}

int main() {
while(Do());
return 0;
}
```