Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 119431    Accepted Submission(s): 27647

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

 

Sample Output

Case 1:14 1 4Case 2:7 1 6

 

思路  ： 用for循环累加  如果和小于0 就清零 然后再改变first  last   

#include <iostream>
#include <stdio.h>
using namespace std;
int main ()
{ int a[100050],i,j,sum1,temp,first,last,t,n,max1;
   scanf("%d",&t);
  for(j=1;j<=t;j++)
   {
       scanf("%d",&n);
       for(i=0;i<n;i++)
        scanf("%d",&a[i]);
        sum1=0,temp=0,last=0,max1=-1001;
       for(i=0;i<n;i++)
       {
           sum1+=a[i];
           if(sum1>max1)
           {
               max1=sum1;
               first=temp;
               last=i;
           }
           if(sum1<0)
           {
               sum1=0;
               temp=i+1;
           }
       }
       printf("Case %d:\n%d %d %d\n",j,max1,first+1,last+1);
       if(j!=t)
        printf("\n");
   }
    return 0;
}