Question

4、 编写一个求1--n内的偶数和的函数，在main()函数中输入n，然后调用该函数求1--n的偶数和并输出

Answer 1

C#代码

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace test
{
class Program
{
static void Main(string[] args)
{
int key = Convert.ToInt32(Console.ReadLine());
int total = 0;
for (int i = 0; i <= key; i++)
{
if (i % 2 == 0)
{
total = total + i;
}
}
Console.WriteLine("{0}以内所有偶数和是{1}",key,total);
}
}
}

追问

有没有C语言的

追答

网上搜了个，你试试
#include "stdio.h"
void main()
{int sun.i,n;
scanf("%d",&n)
sum=0;
for(i=0;i<n;i+=2)
sum=sum+i;
printf("%d",sum);
}

Answer 2

int f(int n)
{
int s=0;
for(int i=1;i<=n;i++)
{
if(i%2==0)
s+=i;
}
return s;
}
其实很简单的

Answer 3

#include<stdio.h>
int sum(int n)
{
int s = 0;
for (int i = 2; i < (n+1); i+=2)
{
s += i;
}
return s;
}

void main()
{
printf("%d\n",sum(1));
printf("%d\n",sum(6));
printf("%d\n",sum(7));
printf("%d\n",sum(8));
}

Answer 4

VC 测试通过
#include <iostream>
using std::cin;
using std::cout;
using std::endl;
int count(int n)
{
int sum=0;
for(int i=1;i<=n;++i)
{
if(i%2==0) //为偶数，则累加
sum+=i;;
}
return sum; //sum即为和
}
int main()
{
int n;
cout << "input a number: " << endl;
cin >> n;
cout << count(n) << endl;
return 0;
}

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