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最新推荐文章于 2022-12-01 17:20:30 发布

原创最新推荐文章于 2022-12-01 17:20:30 发布 · 510 阅读

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本文详细介绍了一种求解子序列最大和的算法，通过实例解释了如何在给定序列中找到具有最大和的连续子序列及其起始和结束位置。文章通过具体代码示例，演示了算法的实现过程。

Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 49213 Accepted Submission(s): 10950

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
//第一个数表示用例数

//a[0]表示字符串个数，a[1]-----a[n-1]为测试用例

//输出最大值，开始终止下标

为Sample Output
Case 1:
14 1 4

Case 2:
7 1 6

思路为，重点在于求子序列最大算法

首先设立一个st，end，max为最大子序列索引，最大值，初始值为1，1，负无穷

其次设立一个sum对数组进行“遍历操作”

如果加了之后sum比max大，就更改max相关

为负了，重置

有人会问？介于之间呢？

不用管，因为虽然没有超过max，但只是不为负，可以继续加，万一加着就超过了呢？没超过大于0也比0好啊

#include "stdafx.h"
#include <stdio.h>
int main(void)
{
	int a, b, sum, max, st, en, e[100000], zo;
	int g = 1;
	scanf_s("%d", &a);
	while (a--)
	{
		st = zo = en = 1;//zo记录sum的开始点
		max = -1001;
		sum = 0;
		scanf_s("%d", &e[0]);
		for (b = 1; b <= e[0]; b++)
		{
			scanf_s("%d", &e[b]);
			sum += e[b];
			if (sum>max)//如果加了之后变大，调整
			{
				max = sum;
				st = zo;
				en = b;
			}
			//如果加了之后在0与max之间呢？也就是加了个e[i]为负，继续加，加到为负了重置，加到超过了调整
			//因为至少还是正数，比变为负数要好
			if (sum<0)//如果为负，重清
			{
				sum = 0;
				zo = b + 1;
			}
		}
		printf("Case %d:\n", g++);
		printf("%d %d %d\n", max, st, en);
		if (a>0)
			printf("\n");
	}
	return 0;
}