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## Project Euler 70–Totient permutation

Wu Yudong    August 20, 2018     欧拉计划   495

Totient permutation

Euler’s Totient function, φ(n) [sometimes called the phi function], is used to determine the number of numbers less than n which are relatively prime to n. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, φ(9)=6.

The number 1 is considered to be relatively prime to every positive number, so φ(1)=1.

Interestingly, φ(87109)=79180, and it can be seen that 87109 is a permutation of 79180.

Find the value of n, 1 < n < 107, for which φ(n) is a permutation of n and the ratio n/φ(n) produces a minimum.

1被认为和任意正整数互质，所以φ(1)=1。

//（Problem 70）Totient permutation
// Completed on Tue, 18 Feb 2014, 11:06
// Language: C11
//
// 版权所有（C）wu yudong
// 博客地址：http://www.wuyudong.com

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#include<stdbool.h>

#define N 10000000

int phi[N];	//数组中储存每个数的欧拉数

int cmp(const void *a, const void *b)
{
return (*(char *)a - *(char *)b);
}

void genPhi(int n)	//求出比n小的每一个数的欧拉数(n-1的)
{
int i, j, pNum = 0;
memset(phi, 0, sizeof(phi));
phi[1] = 1;
for (i = 2; i < n; i++) {
if (!phi[i]) {
for (j = i; j < n; j += i) {
if (!phi[j])
phi[j] = j;
phi[j] = phi[j] / i * (i - 1);
}
}
}
}

int fun(int n)					//计算整数n的位数
{
return (log10(n * 1.0) + 1);
}

bool compare(int n, int m)		//判断两整数是否其中一个是另一个的排列数
{
int i, L1, L2;
char from[10], to[10];
sprintf(from, "%lld", m);
sprintf(to, "%lld", n);
L1 = strlen(from);
L2 = strlen(to);
qsort(from, L1, sizeof(from[0]), cmp);
qsort(to, L2, sizeof(to[0]), cmp);
return !strcmp(from, to);
}

void solve()
{
int i, j, count, k;
double min, t;
min = 10.0;
for (i = 2; i < N; i++) {
if ((fun(i) == fun(phi[i])) && compare(i, phi[i])) {
t = i * 1.0 / phi[i];
if (t < min) {
min = t;
k = i;
}
}
}
printf("%d\n", k);
}

int main()
{
genPhi(N);
solve();
return 0;
}