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

题意：有n个课程，每个课程用不同的时间学习能得到不同的收获。求m天能得到的最大收获。给出的是矩阵形式，比如说第一个样例，是2行2列的矩阵，第一行为第一个课程，学习不同天的收获，第一行第一列那个数为学习第一个课程一天的收获，第二行第二列的那个数为学习第一个课程二天的收获，第二行为第二个课程，收获是多少——（同第一行的解释），这样应该懂了吧，不懂留言哦~

题解：抽象一下，背包容量相当于拥有的时间，然后给你若干组物品（课程），每一组物品里面有着若干个物品（课程时间），一组里面的物品只能选择一个，这么一抽象，标准的分组背包问题，模板题~~~详细见代码：

 

#include <iostream>
#include <cstring>
using namespace std;
const int MAX = 200;
int a[MAX][MAX],dp[MAX];
int main(){
	int n,m;
	while(cin >> n >> m,n+m){
		memset(dp,0,sizeof(dp));
		for (int i = 1; i <= n;i++){
			for (int j = 1; j <= m;j++){
				cin >> a[i][j];
			}
		}
		for (int i = 1; i <= n;i++){
			for (int j = m; j >= 0;j--){//容量循环
				for (int k = 1; k <= m;k++){//个数循环
					if(j-k>=0) dp[j]=max(dp[j],dp[j-k]+a[i][k]);//j-k判断，防止数组越界
				}
			}
		}
		cout << dp[m] << endl;
	}
	return 0;
}