题目:
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
题目链接
直接求欧拉函数值。
AC代码:
//#include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define mp make_pair
#define lowbit(x) (x&(-x))
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
typedef pair<ll,ll> PLL;
const int INF = 0x3f3f3f3f;
const int maxn = 1e6 + 5;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double pi = asin(1.0) * 2;
const double e = 2.718281828459;
bool Finish_read;
template<class T>inline void read(T &x) {
Finish_read = 0;
x = 0;
int f = 1;
char ch = getchar();
while (!isdigit(ch)) {
if (ch == '-') {
f = -1;
}
if (ch == EOF) {
return;
}
ch = getchar();
}
while (isdigit(ch)) {
x = x * 10 + ch - '0';
ch = getchar();
}
x *= f;
Finish_read = 1;
};
ull Phi(ull x) {
ull res = x;
for (int i = 2; i <= int(sqrt(x * 1.0)); ++i) {
if (!(x % i)) {
res = res / i * (i - 1);
while (!(x % i)) {
x /= i;
}
}
}
if (x > 1) {
res = res / x * (x - 1);
}
return res;
}
int main(int argc, char *argv[]) {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
ull n;
while (~scanf("%llu", &n) && n) {
printf("%llu\n", Phi(n));
}
#ifndef ONLINE_JUDGE
fclose(stdin);
fclose(stdout);
system("gedit out.txt");
#endif
return 0;
}