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

这道题主要是输出格式的问题，每两个案例之间用一行空格格开，最后一个案例则不需要。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
const int N=100001;
const int INF=0x3f3f3f3f;
using namespace std;
int a[N];
int main()
{
   int T;
   scanf("%d",&T);
   int counts=1;
   while(T--)
   {
      int n;
      scanf("%d",&n);
      for(int i=1;i<=n;i++)
          scanf("%d",&a[i]);
      int maxn=-1001;
      int sum=0,st=0,en=0;
      int temp=1;
      for(int i=1;i<=n;i++)
      {
            sum+=a[i];
            if(sum>maxn)
            {
               maxn=sum;
               st=temp;
               en=i;
            }
            if(sum<0)
            {
                sum=0;
                temp=i+1;
            }
        }
        printf("Case %d:\n%d %d %d\n",counts++,maxn,st,en);
        if(T)
        printf("\n");
   }
   return 0;
}