HDOJ-1712-ACboy needs your help

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6413    Accepted Submission(s): 3543 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 思路：这是一道分组背包问题，状态转移方程为 dp[l] = max(dp[l-j]+a[j],dp[l]); 需要注意的是要吧 dp数组写在数据数组外面。 #include <stdio.h>#include <string.h>#include <algorithm>int main(){ using namespace std; int n,m; int a[105],dp[105]; int i,j,l; while(~scanf("%d%d",&n,&m) && (n||m)) { memset(dp,0,sizeof(dp)); for(int i=1; i<=n; i++)//几门课 { for(j=1; j<=m; j++)//天数 { scanf("%d",&a[j]); } for(l=m; l>0; l--) { for(j=1; j<=m; j++) { if(l>=j) { dp[l] = max(dp[l] , dp[l-j]+a[j]); } } } } printf("%d\n",dp[m]); } return 0;}