HDOJ-1003: 最大连续和

HDOJ-1003题目如下
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

``````#include <iostream>
#include <algorithm>
using namespace std;
int a[100001];
int main(){
int number;
cin >> number;

for(int i = 1; i <= number; i++){
int max = INT_MIN;
int t,left, right;
cin >> t;
left = right = 1;
int m = 1;
int sum = 0;
for(int j = 1; j <= t; j++){
cin >> a[i];
sum += a[i];
if(sum > max){
max = sum;
right = j;
left = m;
}
if(sum < 0){
sum = 0;
m = j + 1;
}
}
cout << "Case " << i << ":" << endl;
cout << max << " "<< left << " " << right << endl;
if( i < number )
cout << endl;
}
system("pause");
return 0;
}``````