ACboy needs your help
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3215 Accepted Submission(s): 1670
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.
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
Source
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1 //46MS 280K 552 B C++ 2 /* 3 4 题意： 5 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， 6 求花费m天时间能获得的最大价值 7 8 背包： 9 分组背包模板题..将每一天看成一个组，每组只能选一个。 10 11 */ 12 #include<stdio.h> 13 #include<string.h> 14 int a[105][105]; 15 int dp[105]; 16 int Max(int a,int b) 17 { 18 return a>b?a:b; 19 } 20 int main(void) 21 { 22 int n,m; 23 while(scanf("%d%d",&n,&m),n+m) 24 { 25 memset(dp,0,sizeof(dp)); 26 for(int i=1;i<=n;i++) 27 for(int j=1;j<=m;j++) 28 scanf("%d",&a[i][j]); 29 for(int i=1;i<=n;i++) 30 for(int j=m;j>=0;j--) 31 for(int k=1;k<=j;k++) 32 dp[j]=Max(dp[j],dp[j-k]+a[i][k]); 33 printf("%d ",dp[m]); 34 } 35 return 0; 36 }