POJ 2407 Relatives

Description

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.
Input

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.
Output

For each test case there should be single line of output answering the question posed above.
Sample Input

7
12
0

Sample Output

6
4

Source
Waterloo local 2002.07.01
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欧拉函数~

（Euler函数表达通式：euler(x)=x(1-1/p1)(1-1/p2)(1-1/p3)(1-1/p4)…(1-1/pn),其中p1,p2……pn为x的所有素因数，x是不为0的整数。euler(1)=1（唯一和1互质的数就是1本身）。 
     欧拉公式的延伸：一个数的所有质因子之和是euler(n)*n/2。）

这个是欧拉函数的模板~直接套公式的~

int euler(int u)
{
int res=u,a=u;
for(int i=2;i*i<=a;i++)
{
if(a%i==0) res=res/i*(i-1);
while(a%i==0) a/=i;
}
if(a>1) res=res/a*(a-1);
return res;
}

线性筛打欧拉函数表：

phi[1]=1;
for(int i=2;i<=n;i++)
{
if(!b[i]) a[++a[0]]=i,phi[i]=i-1;
for(int j=1;j<=a[0] && i*a[j]<=n;j++)
{
b[a[j]*i]=1;
if(!(i%a[j]))
{
phi[a[j]*i]=phi[i]*a[j];break;
}
phi[a[j]*i]=phi[i]*phi[a[j]];
}
}

这道题的代码：

#include<cstdio>

int t,n;

int euler(int u) { int res=u,a=u; for(int i=2;i*i<=a;i++) { if(a%i==0) res=res/i*(i-1); while(a%i==0) a/=i; } if(a>1) res=res/a*(a-1); return res; }

int main()
{
while(scanf("%d",&n)==1 && n)
{
printf("%d\n",euler(n));
}
return 0;
}