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
题意： 有n个课程，m天时间，接下来n*m的矩阵表示该课程花费1~m天时间所获能得的价值，求花费m天时间能获得的最大价值
背包：
分组背包模板题..将每一天看成一个组，每组只能选一个。
代码实现：
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int max(int a,int b)
{
return a>b?a:b;
}
int main()
{
int n,m,i,j,k,a[][],dp[];
while(cin>>n>>m&&n&&m)
{
for(i=;i<=n;i++)
for(j=;j<=m;j++)
cin>>a[i][j];
memset(dp,,sizeof(dp));
for(i=;i<=n;i++)//分组
for(j=m;j>;j--)//容量
for(k=;k<=j;k++)//第i组数据
dp[j]=max(dp[j],dp[j-k]+a[i][k]);
cout<<dp[m]<<endl;
}
return ;
}