1003 局部最值问题

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

#include<stdio.h>
int i,cas,j,k,t,max,s,e,n,x;
main()
{
    while(scanf("%d",&cas)!=EOF)
    {
        for(i=0;i<cas;i++)
        {
            if(i)printf("/n");
            printf("Case %d:/n",i+1);
            scanf("%d",&n);
            max=-9999;
            for(j=k=1,t=0;j<=n;j++)
            {
                scanf("%d",&x);
                t+=x;
                if(t>max)  {max=t;s=k;e=j;}
                if(t<0)      {k=j+1;t=0;}
            }
            printf("%d %d %d/n",max,s,e);
        }
    }
}