C语言程序 编一函数，计算正整数1～n中的奇数之和和偶数之和，要求函数的原型为long sum(int x) 。

Question

如题所述

Answer 1

long sum_ji(int x) //奇数之和
{
if（x % 2 == 0）{
x--;
}
return (long)x * (long)x / 4;
}

long sum_ou(int x) //偶数之和
{
if(x % 2 != 0){
x--;
}
return (long)(x+2) * (long)x /4;
}本回答被提问者和网友采纳

Answer 2

#include<stdio.h>
long sum(int x)
{
int sum,i;
for (sum=0,i=1;i<=x;i+=2)
{sum=sum+i;}
printf("奇数之和为:%d\n",sum);
for (sum=0,i=0;i<=x;i+=2)
{sum=sum+i;}
return sum;
}

int main()
{
int a;
printf("please input a int number:");
scanf("%d",&a);
printf("偶数之和为:%d\n",sum(a));
return 0;
}
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