HDU 1712 ACboy needs your help

分组背包
HDU 1712
Problem Description
ACboy has N courses this term, and he plans to spend at most M days on study.
Of course,the profit(利润) 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(取…最大值) the profit?


Input
The input consists of multiple data sets.
A data set starts with a line containing two positive(积极的) integers(整数) N and M, 
N is the number of courses, M is the days ACboy has.
Next follow a matrix(矩阵) A[i][j], (1<=i<=N<=100,1<=j<=M<=100).
A[i][j] indicates(表明) 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(输出) 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：
一个学生用M天的时间复习N门课程，每门课程花费不同的天数，有不同的收获。问如何安排这M天，使得收获最大。
思路：
分组背包（背包九讲）

for 所有的组k

    for v=V..0

        for 所有的i属于组k

            f[v]=max{f[v],f[v-c[i]]+w[i]}

#include <stdio.h>#include <iostream>#include <cstring>using namespace std;int max(int x,int y){    return x>y?x:y;}int main(){    int n,K,i,j;    int dp[105],a[105];    while(cin>>n>>K)    {        if(n+K==0)        break;        memset(dp,0,sizeof(dp));        for(i=1;i<=n;i++)        {            for(j=1;j<=K;j++)            {                cin>>a[j];            }            for(j=K;j>0;j--)            {                for(int k=j;k>0;k--)                    dp[j]=max(dp[j],dp[j-k]+a[k]);            }        }        cout<<dp[K]<<endl;    }    return 0;}