ACboy needs your help （动态规划背包）

ACboy has N courses this term, and he plans to spend at most M days on study.Of course,the profit he will gain from different course depending on the days he spend on it.How to arrange the M days for the N courses to maximize the profit? 

Input

The input consists of multiple data sets. A data set starts with a line containing two positive integers N and M, N is the number of courses, M is the days ACboy has. 
Next follow a matrix A[i][j], (1<=i<=N<=100,1<=j<=M<=100).A[i][j] indicates if ACboy spend j days on ith course he will get profit of value A[i][j]. 
N = 0 and M = 0 ends the input. 

Output

For each data set, your program should output a line which contains the number of the max profit ACboy will gain. 

Sample Input

2 2
1 2
1 3
2 2
2 1
2 1
2 3
3 2 1
3 2 1
0 0

Sample Output

3
4
6

题解：动态规划背包问题，三重循环解即可

代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;

int main()
{
    int n,m;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
	    if(n==0||m==0)
		{
		break;	
		}	
		long long int dp[105];
		int a[105][105];
		memset(dp,0,sizeof(dp));
        memset(a,0,sizeof(a));

		for(int t=1;t<=n;t++)
		{
			for(int j=1;j<=m;j++)
			{
				scanf("%d",&a[t][j]);
			}
		}
		for(int t=1;t<=n;t++)
		{
			for(int j=m;j>=0;j--)
			{
				for(int k=0;k<=j;k++)
				{
					dp[j]=max(dp[j],dp[j-k]+a[t][k]);
				}
			}
		}
		printf("%lld\n",dp[m]);
	}	
  return 0;
}