2^x mod n = 1
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9231 Accepted Submission(s): 2837
Problem Description
Give a number n, find the minimum x(x>0) that satisfies 2^x mod n = 1.
Input
One positive integer on each line, the value of n.
Output
If the minimum x exists, print a line with 2^x mod n = 1.
Print 2^? mod n = 1 otherwise.
You should replace x and n with specific numbers.
思路：
1. 当n为偶数时，bn + 1（b为整数）是奇数，而2^x是偶数，故 2^x mod n = 1不可能成立；
2. 当n等于1时，不能成立
3. 当n为非1的奇数时，n和2互质，由欧拉定理：若a，n为正整数，且两者互素，则a^phi(n) mod n = 1，其中phi(n)是n的欧拉函数。知2^phi(n) mod n = 1.因此phi(n)必是符合要求的x，但phi(n)未必是最小的，遍历小于其的正整数，逐一试验即可，计算2^x mod n时用快速幂乘。
AC Code：
#include <iostream>
#include <vector>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std; //计算n的欧拉函数
int Eular(int n)
{
int res = , i;
for (i = ; i * i <= n; i++){
if (n % i == ){
n /= i;
res *= (i - );
while (n % i == ){
n /= i;
res *= i;
}
}
}
if (n > ) res *= (n - );
return res;
} //快速幂乘计算2^b % n
int myPow(int b, int n)
{
if(b == ) return ;
long long c = myPow(b >> , n);
c = (c * c) % n;
if(b & ) c = ( * c) % n;
return c;
} int main()
{
int n, x;
bool ok;
while(scanf("%d", &n) != EOF){
ok = ;
if((n & ) && (n - )){
ok = ;
int phi = Eular(n);
for(x = ; x < phi; x++){
if(myPow(x, n) == ) break;
}
}
if(ok) printf("2^%d mod %d = 1\n", x, n);
else printf("2^%? mod %d = 1\n", n);
}
return ;
}