New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Jump to bottom

王天锐 #27

Open

wangtianrui opened this issue Oct 30, 2016 · 0 comments

Member

wangtianrui commented Oct 30, 2016

迭代法输出第N个斐波那契数

#include<stdio.h>
int f( int x )
{
    int y ;
    if( x == 1 || x == 2)
    {
        return ( 1 ) ;
    }
    else
    {
        y = f( x - 1 ) + f( x - 2 ) ;
        return ( y ) ;
    }
}

int main ()
{
    int  shu , n ;
    printf("你想求的第几个斐波那契数？");
    scanf("%d", &n );
    shu = f ( n ) ;
    printf("%d", shu ) ;
    return 0 ;
}

Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment