OJ1003

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;
}