HDU 1712 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

大致题意：一个人有n门课，准备在m天内学习这些课，然后给你一个n*m的矩阵，a[i][j]表示第i门课学习j天的收益，一天只能学习一门课，问最大收益多少。

思路：比较裸的分组背包
代码如下

//#include<bits/stdc++.h>#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<map>using namespace std; #define LL long long intint a[105][105];int dp[105];int main(){    int n,m;    while(scanf("%d%d",&n,&m)!=EOF)    {        if(!(n+m)) break;        memset(dp,0,sizeof dp);        for(int i=1;i<=n;i++)        for(int j=1;j<=m;j++)        scanf("%d",&a[i][j]);        for(int i=1;i<=n;i++)//i组        for(int j=m;j>=0;j--)        for(int k=1;k<=m;k++)            if(j-k>=0)            {                dp[j]=max(dp[j],dp[j-k]+a[i][k]);            }        printf("%d\n",dp[m]);    }}