Making Money

FJ has gone into the curio business, buying and selling knickknacks like cow Christmas tree ornaments. He knows he will sell every single curio he can stock from a catalog of N (1 <= N <= 100) different cow curios, and he can buy as many of each kind of curio as his heart desires. He has only M (1 <= M <= 100,000) money to invest but wants to maximize his profit (which has a slightly unusual definition) at the end of his first year in business. 
Curio type i costs C_i (1 <= C_i <= 100,000) money to purchase and yields R_i (1 <= R_i <= 100,000) revenue for each curio sold (a profit of R_i-C_i). FJ can mix and match the curios he sells in any way he wishes. He need not spend all his money when purchasing curios. 
What is the greatest amount of total profit (profit = initial_cash - all_costs + all_sales) FJ can have at the end of his first year? This number is guaranteed to be less than 1,000,000,000. 
Consider the situation when FJ has just 3 kinds of curios and starts with M=17. Below are the cost and revenue numbers for each curio: 

             Curio     Cost     Revenue

               #        C_i       R_i

               1         2         4

               2         5         6

               3         3         7

输入

#include <bits/stdc++.h>
using namespace std;
const int Max=105;
#define INF 0x3f3f3f3f
int dp[100005],a[Max],v[Max];
int main()
{
    int n,m,k;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        int b;
        scanf("%d%d",&a[i],&b);
        v[i]=b-a[i];
    } 
    for(int i=1;i<=n;i++)
    {
        for(int j=a[i];j<=m;j++)
        dp[j]=max(dp[j],dp[j-a[i]]+v[i]);
    }  //01背包 
    int Max=0;
    for(int i=1;i<=m;i++)
    Max=max(Max,m-i+dp[i]); //  剩下的钱+收入 
    cout<<Max<<endl;
}