ACboy needs your help（分组背包）

Problem Description
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

伪代码：for 1...i组
for V....0
for 所有的k属于i组
dp[j]=max(dp[j],dp[j-v[i]]+w[i]);

#include<stdio.h>#include<string.h>#define Max 105int w[Max][Max],dp[Max];int m,n;int main(){//freopen("b.txt","r",stdin);int i,j,k;while(scanf("%d %d",&m,&n)==2){if(m+n==0) break;for(i=1;i<=m;i++)for(j=1;j<=n;j++)scanf("%d",&w[i][j]);memset(dp,0,sizeof(dp));for(i=1;i<=m;i++)for(j=n;j>=0;j--){for(k=1;k<=n;k++)if(j>=k) dp[j]=dp[j]>dp[j-k]+w[i][k]?dp[j]:dp[j-k]+w[i][k];}printf("%d\n",dp[n]);}return 0;}