（Problem 70）Totient permutation

Euler's Totient function, φ(n) [sometimes called the phi function], is used to determine the number of positive numbers less than or equal to 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.

题目大意：

欧拉函数φ(n)（有时也叫做phi函数）可以用来计算小于等于n 的数字中与n互质的数字的个数。例如，因为1,2,4,5,7,8全部小于9并且与9互质，所以φ(9)=6。

数字1被认为与每个正整数互质，所以 φ(1)=1。

有趣的是，φ(87109)=79180，可以看出87109是79180的一个排列。

对于1  n  10^7，并且φ(n)是 n 的一个排列的那些 n 中，使得 n/φ(n) 取到最小的 n 是多少？

//（Problem 70）Totient permutation
// Completed on Tue, 18 Feb 2014, 11:06
// Language: C11
//
// 版权所有（C）acutus   (mail: acutus@126.com) 
// 博客地址：http://www.cnblogs.com/acutus/

#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
", k);
}

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

Answer:

8319823