Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 98479    Accepted Submission(s): 22665

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


注意presentation error
最后的一个判断！！！
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<vector>
#include <math.h>
#define INF -100000000
using namespace std;
int sum[100010];
int a[100010];
int main()
{
    int t;
    while(scanf("%d",&t)==1)
    {
        for(int j = 0;j<t;++j)
        {
            int k;
            scanf("%d",&k);
        memset(sum,0,sizeof(sum));
        int maxn  = INF;
        int l = 1,r = 1;
        int L,R;
        if(k==0)
        break;
        for(int i = 1;i<=k;++i)
         scanf("%d",&a[i]);
        for(int i = 1;i<=k;++i)
        {
         if(a[i]>sum[i-1]+a[i])
           {
               sum[i] = a[i];
               l = i;
               r = i;
           }
          else
          {
             sum[i] = sum[i-1]+a[i];
             r = i;
          }
         if(sum[i]>maxn)
         {
             L = l;
             R = r;
             maxn = sum[i];
         }
        }
        //if(maxn < 0)
        //printf("Case %d:\n%d %d %d\n",j+1,0,a[1],a[k]);
        //else
        printf("Case %d:\n%d %d %d\n",j+1,maxn,L,R);
        if(j!=t-1)
        printf("\n");
    }
    }
    return 0;
}