hdu 1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 100827    Accepted Submission(s): 23197

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

 

Author

Ignatius.L

//用sum和max_sun来记录临时最大和最终最大:
//1.当sum < 0时，重新计算，并记录此时临时下标，用于标记开始坐标。
//2.当sum > max_sum 时，记录开始下标和结束下标。
#include <stdio.h>
int main()
{
	int t, n, a, sum, max_sum, start, temp, end, i, k = 0;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d", &n);
		sum = -1;
		max_sum = -1001;
		for(i = 1; i <= n; i++)
		{
			scanf("%d", &a);
			if(sum < 0)
			{
				sum = a;
				temp = i;
			}
			else
				sum += a;
			if(sum > max_sum)
			{
				max_sum = sum;
				start = temp;
				end = i;
			}
		}
		k++;
		printf("Case %d:\n%d %d %d\n", k, max_sum, start, end);
		if(t)
			printf("\n");
	}
	return 0;
}