hdoj 1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 115460    Accepted Submission(s): 26774

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>

int main()
{	
	int n, star, end, i, j, t, x, temp, now, sum,count = 1;
	scanf("%d", &t);
	for(i = 1; i <= t; i++)
	{
		scanf("%d %d", &n, &temp);
		star = end = x = 1;
		now = sum = temp;
		for(j = 2; j <= n; j++)
		{
			scanf("%d", &temp);
			if(now + temp < temp)
			{
				now = temp;
				x = j;
			}
			else
			{
				now += temp;
			}
			if(now > sum)
			{
				sum = now;
				star = x;
				end = j;
			}
		}
		printf("Case %d:\n%d %d %d\n", count++, sum, star, end);
		if(i != t)
		{
			printf("\n");
		}
	}
	return 0; 
}

 

这题代码是看了别人的。。很简单的一道题  

总体思路就是先把第一个数当做最大子序列   初始化各个变量

然后输入下一个数,将其加入到子序列中，如果now+temp<temp 就说明

这一串子序列加起来还没有temp大 所以now=temp  此时更改起始位置的下标

否则的话就说明加入后子序列的值变大  所以将其加入到子序列中去。

如果加入后的值大于max则更新max的值，

   各变量的含义如下

 

star          子序列的起始位置

end          子序列的最后一个位置

now          加入一个数后的当前最大值

max          保存子序列的最大值

x               子序列的起始位置