1874: Relatives
| Status | In/Out | TIME Limit | MEMORY Limit | Submit Times | Solved Users | JUDGE TYPE |
|---|---|---|---|---|---|---|
| stdin/stdout | 3s | 8192K | 415 | 177 | Standard |
Given
n, a positive integer, how many positive integers less than
n are relatively prime to
n? Two integers
a and
b are relatively prime if there are no integers
x > 1, y > 0, z > 0 such that
a = xy and
b = xz.
There are several test cases. For each test case, standard input contains a line with n <= 1,000,000,000. A line containing 0 follows the last case.
For each test case there should be single line of output answering the question posed above.
Sample Input
7 12 0
Output for Sample Input
6 4
#include<stdio.h>
int main()
{
int n,k;
while(scanf("%d",&n)&&n)
{
int b=n;k=2;
int res=n;
while(k<=b)
{
if(b%k==0)
{
res-=res/k;
b/=k;
}
while(b%k==0) b/=k;
if(b==1) break;
k++;
}
if(b==n) {printf("%d/n",n-1);continue;}
//if(b>1&&b<n) res-=res/k;
printf("%d/n",res);
}
return 0;
}
本文详细介绍了如何计算小于给定整数的所有相对质数的数量,并提供了代码实现。

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