HDU-1003 Max Sum

Given a sequence a11,a22,a33......ann, 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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int num[100005];
int main()
{
 int t, cases, sum, mmax;
 scanf("%d", &t);
 for(int k = 1; k <= t; k++)
 {
     sum = 0;
     mmax = -1005;
     int si = 1, sj = 1;
     scanf("%d", &cases);
     memset(num, 0, sizeof(num));
     for(int i = 1; i <= cases; i++)
        scanf("%d", &num[i]);
        int i = 1;
     for(int j = 1; j <= cases; j++)
     {
         sum += num[j];
         if(mmax < sum)  {mmax = sum; si = i; sj = j;} //更新最大值
         if(sum < 0) {i = j + 1;  sum = 0;} //总和小于0则舍弃
     }
     if(k - 1) printf("\n");
     printf("Case %d:\n", k);
     printf("%d %d %d\n", mmax, si, sj);
 }
    return 0;
}