C Exercise Example 76
100 cases of classic C language
Title: Write a function that input n is an even number, call the function evaluation 1/2 + 1/4 + ... + 1 / n, when the input n is odd, the function is called 1/1 + 1/3 + .. . + 1 / n (use the pointer function).
Program Analysis: None.
Source Code:
// Created by www.w3big.com on 15/11/9.
// Copyright © 2015年 本教程. All rights reserved.
//
#include<stdio.h>
#include<stdlib.h>
double evenumber(int n);
double oddnumber(int n);
int main()
{
int n;
double r;
double (*pfunc)(int);
printf("请输入一个数字：");
scanf("%d",&n);
if(n%2==0) pfunc=evenumber;
else pfunc=oddnumber;
r=(*pfunc)(n);
printf("%lf\n",r);
system("pause");
return 0;
}
double evenumber(int n)
{
double s=0,a=0;
int i;
for(i=2;i<=n;i+=2)
{
a=(double)1/i;
s+=a;
}
return s;
}
double oddnumber(int n)
{
double s=0,a=0;
int i;
for(i=1;i<=n;i+=2)
{
a=(double)1/i;
s+=a;
}
return s;
}
Run the above example output is:
请输入一个数字：2 0.500000