HDU1712 ACboy needs your help

HDU 水题 背包问题 分组背包问题 注意 这个

            for (int j=1;j<=M;j++)
                scanf("%d",&A[j]);

不能写成这个

            for (int j=0;j<M;j++)
                scanf("%d",&A[j]);

题目及AC代码 如下

ACboy needs your help

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3626    Accepted Submission(s): 1880

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.

 

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

 

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

const int max_n=110;

int max(int a,int b)
{
    return a>b?a:b;
}

int main(void)
{
    int A[max_n];
    int dp[max_n];
    int M,N;
    while(scanf("%d%d",&N,&M)&& (N!=0 && M!=0))
    {
        memset(dp,0,sizeof(dp));
        memset(A,0,sizeof(A));
        for(int i=0;i<N;i++)
        {
            for (int j=1;j<=M;j++)
                scanf("%d",&A[j]);

            for(int vi=M;vi>=0;vi--)
            {
                for(int k=0;k<=vi;k++)
                    dp[vi]=max(dp[vi],dp[vi-k]+A[k]);
            }
        }
        printf("%d
",dp[M]);
    }


    return 0;
}