ACboy needs your help HDU

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 21 21 32 22 12 12 33 2 13 2 10 0

Sample Output

346

分组背包与０１背包基本是一样的，无非是把一组中的元素枚举一遍，看一下能不能装进去，这样枚举的话也保证了每一组最多装了一个

#include<bits/stdc++.h>using namespace std;int dp[120];int gra[120][120];int main(){    int n,m;    while(scanf("%d%d",&n,&m)!=EOF&&(m+n))    {        for(int i=1; i<=n; i++)            for(int j=1; j<=m; j++)                scanf("%d",&gra[i][j]);        memset(dp,0,sizeof(dp));        for(int i=1; i<=n; i++)        {            for(int j=m; j>=1; j--)            {                for(int k=1; k<=j; k++)                    dp[j]=max(dp[j],dp[j-k]+gra[i][k]);            }        }        printf("%d\n",dp[m]);    }}