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标题:指点一下。。。。
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既生瑜何生亮
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指点一下。。。。
#include<stdio.h>
int main()
{
     long int a[100002],i,t,m,n,h,j,sum;
    scanf("%d",&m);
    for(i=1;i<=m;i++)
    {   
        sum = 0;
        scanf("%d",&n);
        for(j=1;j<=n;j++)
            scanf("%d",&a[j]);
        t = a[1]; h = 1;
        for(j=1;j<= n;j++)
        {
            sum+=a[j];
            if(sum>t)
            {
                t = sum;
                h = j ;
            }
        }
        printf("Case %d:\n",i);
        printf("%d %d %d\n",t,1,h);
        if(i!=m)
            printf("\n");
    }
    return 0;
}Problem Description
Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

 

Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).

 

Output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.

 

Sample Input
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5
 

Sample Output
Case 1:
14 1 4

Case 2:
7 1 6
 http://acm.hdu.
我想问一下。我的代码有什么疏漏之处。。

 

        
搜索更多相关主题的帖子: include 
2012-07-11 21:50
既生瑜何生亮
Rank: 1
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想了一下,好像略有所懂。求的那段子序列不一定是从第一个开始的。。。
2012-07-11 22:03
快速回复:指点一下。。。。
数据加载中...
 
   



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