1003

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
  
  
  
  

   
   这道题运用了动规思想，要注意的是这道题多了一个位置的记录，所以起点额选择会更新，因此要定义一个变量去记录起点的位置，而且还要注意输出的格式。
  
  
  
  

   
   
   
   
#include<iostream>
#include<cstdio>
using namespace std;

int main()
{
    int n;
    int a[100010];
    scanf("%d",&n);
    int k=0,m,maxn,temp,position,begining,ending,i;
    while(n--)
    {
        k++;
        scanf("%d",&m);
        for(int i=0;i<m;i++)
        {
            scanf("%d",&a[i]);
        }
        maxn=temp=a[0];
        position=begining=ending=0;
        for( i=1;i<m;i++)
        {
            if(temp+a[i]<a[i])
            {
                temp=a[i];
                position=i;
            }
            else
            {
                temp+=a[i];
            }
            if(temp>maxn)
            {
                maxn=temp;
                begining=position;
                ending=i;
            }

        }
        printf("Case %d:\n%d %d %d\n",k,maxn,begining+1,ending+1);
        if(n)
            printf("\n");

    }
    return 0;
}