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Time Limit:1000MS
Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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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，现在花M天来学习这些课程，学习每个课程花的天数所得到的价值不同，求M天怎么分配学习才能得到的价值最大。
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
int main()
{
int n,m,i,j,k;
int f[110][110];
int dp[110];
while(~scanf("%d %d",&n,&m))
{
if(n==0&&m==0)
break;
memset(dp,0,sizeof(dp));
for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
scanf("%d",&f[i][j]);
for(i=1;i<=n;i++)//分组数
for(j=m;j>=1;j--)//容量体积
for(k=1;k<=j;k++)//属于i组的k
dp[j]=max(dp[j],dp[j-k]+f[i][k]);
printf("%d\n",dp[m]);
}
return 0;
}