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 W_i (1 ≤ W_i ≤ 400), a ‘desirability‘ factor D_i (1 ≤ D_i ≤ 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: W_i and D_i

* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints

#include<iostream>
#include<string.h>

using namespace std;

int max(int a,int b){
return a>b?a:b;
}
int main(){
    int wi,di;
    int dp[15000];

    int N,M;

    memset(dp,0,sizeof(dp));
    cin>>N>>M;
    for(int i=1;i<=N;i++){
        cin>>wi>>di;//从输入开始判断能否放入这个珠子

        for(int j=M;j>0;j--){//同样通过数组记录最大值
            if(j>=wi)
                dp[j]=max(dp[j],dp[j-wi]+di);//放完后的价值和未放之前的价值与放入的价值之和相比较，大的那方更新数组
        }
    }
    cout<<dp[M]<<endl;
}