P1048 [noip2005 popularization group] Drug collection

Title Description

Chen Chen is a gifted child , His dream is to become the greatest doctor in the world . So , He wants to learn from the most prestigious doctor in the neighborhood . In order to judge his qualifications , Gave him a problem . The doctor took him to a cave full of herbs and said ：“ children , There are some different herbs in this cave , It takes time to pick each one , Each has its own value . I'll give you a period of time , In the meantime , You can pick some herbs . If you are a smart kid , You should be able to maximize the total value of the herbs you pick .”

If you are Chen Chen , Can you finish the task ？

Input format

The first line has  2  It's an integer  T（1≤T≤1000） and  M（1≤M≤100）, Separated by a space ,T  Represents the total time that can be used to collect medicine ,M  Represents the number of herbs in the cave .

Next M Each line consists of two in 1  To 100 Between （ Include 1  and 100） The integer of , It means the time of picking a herb and the value of the herb .

Output format

Output the maximum total value of herbs that can be collected within the specified time .

I/o sample

Input #1

70 3
71 100
69 1
1 2

Output #1

3

explain / Tips

【 Data range 】

• about 30%  The data of ,0M≤10;
• For all the data ,0M≤100.

【 Title source 】

NOIP 2005 Third question of popularization group

#include<bits/stdc++.h>
using namespace std;
int T,M;
int v[10001],w[10001];
int dp[10001][10001];

int main()
{
    while(cin>>T>>M){
        for(int i=1;i<=M;i++)
            cin>>w[i]>>v[i];
        for(int i=1;i<=M;i++)
            for(int j=1;j<=T;j++){
                if(w[i]>j) 
                    dp[i][j]=dp[i-1][j];
                else
                    dp[i][j]=max(dp[i-1][j],dp[i-1][j-w[i]]+v[i]);
            }
        cout<<dp[M][T]<<endl;
    }
}