Max Sum http://acm.hdu.edu.cn/showproblem.php?pid=1003-优快云博客

本文介绍了一个经典的编程问题——MaxSum，旨在找出数列中子序列的最大和及其位置。通过示例详细解释了如何通过一次遍历算法高效解决该问题，并提供了完整的C语言实现代码。

Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 113459 Accepted Submission(s): 26237

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

#include<stdio.h>
# define M 100000
int main()
{
	int i,n;
	int a[M+5];
	scanf("%d",&n);
	for(i=1;i<=n;i++)
	{
		int j,k,q=1,m,max=-1001,s=0,e=0,sum=0;
		scanf("%d",&m);
		for(j=0;j<m;j++)
		{
			scanf("%d",&a[j]);
			sum+=a[j];
			if(max<sum)
			{
				max=sum;
				s=q;
				e=j+1;
			}
			if(sum<0)
			{
				sum=0;
				q=j+2;
			}
		}
		printf("Case %d:\n%d %d %d\n",i,max,s,e);
		if(i!=n)
			printf("\n");
	}
	return 0;
}

转载于:https://www.cnblogs.com/wangyouxuan/p/3243294.html