| Time Limit: 1000MS | Memory Limit: 32768KB | 64bit IO Format: %I64d & %I64u |
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?
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.
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
/*
Author: 2486
Memory: 1456 KB Time: 93 MS
Language: G++ Result: Accepted
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long LL;
const int maxn=100+5;
int dp[maxn],a[maxn][maxn];
int n,m;
int main() {
while(~scanf("%d%d",&n,&m),n&&m) {
for(int i=1; i<=n; i++) {
for(int j=1; j<=m; j++) {
scanf("%d",&a[i][j]);
}
}
memset(dp,0,sizeof(dp));
int Max=0;
for(int i=1; i<=n; i++) {
for(int k=m; k>=0; k--) {
for(int j=1; j<=m; j++) {
if(k-j<0)break;
dp[k]=max(dp[k],dp[k-j]+a[i][j]);
}
}
}
printf("%d\n",dp[m]);
}
return 0;
}