软件工程第三次作业（2019）

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;
}

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