Max Sum

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

题目大意：给出一组数，求出最大的子序列

代码如下：

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
int main()
{
    int i,cs,testnum;
    int n,number,sum,start,end,temp,max;
    scanf("%d",&testnum);
    for(cs=1;cs<=testnum;cs++)
    {
        max=-1010;
        sum=0;
        temp=1;
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {
            scanf("%d",&number);
            sum+=number;
            if(sum>max)
            {
                max=sum;
                start=temp;
                end=i;
            }
            if(sum<0)
            {
                sum=0;
                temp=i+1;
            }
        }
        printf("Case %d:\n%d %d %d\n",cs,max,start,end);
        if(cs!=testnum)
        printf("\n");
    }
    return 0;
}