最大连续子段和

题目：HDU 1003 Max Sum
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 <stdio.h>
#include <string.h>
#include <stdlib.h>

/*最大连续最大子段和 
* 参数：
* a[] 输入数组 
* n   数组元素个数
* start, end 返回的最大子段和的起始和结束位置 
*/
int maxsum(int a[], int n, int &start, int &end)
{
	int tmp = a[0];
	int max = a[0];
	int i, tmpstart = 0, tmpend = 0;
	start = 1;
	end = 1;
	for(i = 1; i < n; i++)
	{
		if(tmp >= 0) //可以是>0 也可以是>=0，看怎么要求 
		{
			tmp += a[i];
			tmpend = i;	
		}	
		else
		{
			tmp = a[i];
			tmpstart = i;
			tmpend = i;	
		}
		if(max < tmp)
		{
			start = tmpstart + 1;
			end = tmpend + 1;
			max = tmp;	
		}
	}	
	return max;
}
int main()
{
	int tc, n, tmp;
	int start, end;
	int max;
	int i, j; 
	int a[100001];
	scanf("%d", &tc);
	for(j = 0; j < tc; j++)
	{
		scanf("%d", &n);

		for( i = 0; i < n; i++)
			scanf("%d", &a[i]);

		max = maxsum(a, n, start, end);
		printf("Case %d:\n", j + 1);
		printf("%d %d %d\n", max, start, end);
		if(j != tc - 1)
		printf("\n");
	}
	
	return 0;	
}