（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 107，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: [email protected]) // 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\n", k); } int main() { genPhi(N); solve(); return 0; }
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Answer:
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8319823 |