1.题目要求

题目:最大连续子数组和(最大子段和)
问题: 给定n个整数(可能为负数)组成的序列a[1],a[2],a[3],…,a[n],求该序列如a[i]+a[i+1]+…+a[j]的子段和的最大值。当所给的整数均为负数时定义子段和为0,依此定义,所求的最优值为: Max{0,a[i]+a[i+1]+…+a[j]},1<=i<=j<=n
例如,当(a[1],a[2],a[3],a[4],a[5],a[6])=(-2,11,-4,13,-5,-2)时,最大子段和为20。
-- 引用自《百度百科》

2.实现代码


#include <iostream>
using namespace std;

int sum(int a[] ,int count)
{
	int b[100];
	int i;
	int max;
	b[0] = a[0];
	max = b[0];
	for (i = 1; i < count; i++)
	{
		if (b[i - 1] > 0)
			b[i] = b[i - 1] + a[i];
		else
			b[i] = a[i];
		if (b[i] > max)
			max = b[i];
	}
	return max;
}

int main()
{
	int count;
	int a[100];
	int i;
	int max;
	cin >>count;
	for (i = 0; i < count; i++)
	{
		cin >> a[i];
	}
	max = sum(a, count);
	cout << max;
	return 0;
}

代码地址

3.测试代码

程序流程图

条件组合 执行路径
b[i-1]>0,b[i]>max abdef
b[i-1]<=0,b[i]>max acdef
b[i-1]>0,b[i]<=max abdf
b[i-1]<=0,b[i]<=max acdf

测试用例
a[]={1,5,9},max=15
a[]={-1,5,-1},max=5
a[]={-8,-2,-5,8},max=8
a[]={ -2,11,-4,13,-5,-2},max=20

            TEST_METHOD(TestMethod1)
	{
		int max, num[3] = { 1,5,9 };
		max = sum(num, 3);
		Assert::AreEqual(max, 15);
	}

	TEST_METHOD(TestMethod2)
	{
		int max, num[3] = { -1,5,-1 };
		max = sum(num, 3);
		Assert::AreEqual(max, 5);
	}
	TEST_METHOD(TestMethod3)
	{
		int max, num[4] = { -8,-2,-5,8 };
		max = sum(num, 4);
		Assert::AreEqual(max, 8);
	}
	TEST_METHOD(TestMethod4)
	{
		int max, num[6] = { -2,11,-4,13,-5,-2 };
		max = sum(num, 6);
		Assert::AreEqual(max, 20);
	}

4.测试结果

 posted on 2019-04-21 21:36  6886641  阅读(105)  评论(0编辑  收藏  举报