NYOJ 479 Coprimes

Coprimes

时间限制：500 ms  |  内存限制：4096 KB

难度：1

描述

For given integer N (1<=N<=10⁴) find amount of positive numbers not greater than N that coprime with N. Let us call two positive integers (say, A and B, for example) coprime if (and only if) their greatest common divisor is 1. (i.e. A and B are coprime iff gcd(A,B) = 1).

输入

Input file contains integer N.（ 注意：程序以文件结束符“EOF”结束输入。）

输出

Write answer in output file. 每个数字占一行。

样例输入

9

样例输出

6

 

思路：题目比较简单，就是求n以内与n的最大公约数为1的数的个数。

 

#include <stdio.h>int gcd(int m, int n){int r = n;while (r != 0){r = m % n;m = n;n = r;}return m;}int main(){int i,n;while (scanf("%d",&n) != EOF){int total = 0;for (i=1; i<n; i++){if (gcd(i,n) == 1){total++;}}printf("%d\n",total);}return 0;}

 

如果一个质数p整除n，那么当前就有(1-1/p)个数与它的最大公约数不被p整除，累乘即可得出结果。

#include <stdio.h>int main(void){    int i,n,total;    while (scanf("%d", &n) != EOF){total = n;for (i=2; i<=n; i++){if (n % i == 0){total = total / i * (i-1);}while (n % i == 0){n /= i;}}printf("%d\n",total);}    return 0;}