2^x mod n = 1

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19442 Accepted Submission(s): 6093

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.

Sample Input

Sample Output

2^? mod 2 = 1

2^4 mod 5 = 1

Author

MA, Xiao

Source

ZOJ Monthly, February 2003

欧拉定理：a^x 1(mod m),那么 x = euler(m);

当gcd(n,x) > 1 或者 n ==1 的时候，不符合；否则 x = euler(n)的满足这个的最小因数；

#include <bits/stdc++.h>
using namespace std;
#define clr(a) memset(a,0,sizeof(a))
#define line cout<<"-----------------"<<endl;

typedef long long ll;
const int maxn = 1e5+10;
const int MAXN = 1e6+10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9+7;
const int N = 1010;

ll Euler(ll n){//欧拉函数模板
    ll ans = 1;
    for(int i = 2; i * i <= n; i++){
        if(n % i == 0){
            n /= i;
            ans *= (i - 1);
            while(n % i == 0){
                n /= i;
                ans *= i;
            }
        }
    }
    if(n > 1) ans *= (n-1);
    return ans ;
}
ll Mul(ll x , ll y , ll mod){//快速幂取模模板
    ll ans = 0;
    while(y){
        if(y & 1){
            ans = (ans + x) % mod;
        }
        x = x * 2 % mod;
        y >>= 1;
    }
    return ans ;
}
ll pow_mod(ll x , ll n , ll mod ){
    ll ans = 1;
    ll t = x % mod ;
    while(n) {
        if(n & 1){
            ans = Mul(ans , t, mod);
        }
        t = Mul(t , t , mod);
        n >>= 1 ;
    }
    return ans ;
}

int main(){
    ll n;
    while(scanf("%lld",&n)!=EOF){
        if(n % 2 == 0 || n == 1){
            printf("2^? mod %lld = 1\n",n);
            continue;
        }
        ll m = Euler(n);
        ll ans = 1e18;
        for(ll i = 1 ; i * i <= m ; i++){
            if(m % i == 0){
                if(pow_mod(2 , i , n) == 1) 
                    ans = min(ans , i);
                if(pow_mod(2 , m/i , n) == 1)
                    ans = min(ans , m/i);
            }
        }
        printf("2^%lld mod %lld = 1\n", ans , n);
    }
    return 0;
}

#include <bits/stdc++.h>
using namespace std;
#define clr(a) memset(a,0,sizeof(a))
#define line cout<<"-----------------"<<endl;
 
typedef long long ll;
const int maxn = 1e5+10;
const int MAXN = 1e6+10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9+7;
const int N = 1010;
 
vector<ll>ve;

void get_prime(ll n){
	for(ll i = 1 ;i * i <= n;i++){
		if(n % i == 0){
			ve.push_back(i);
			if(i * i != n)
				ve.push_back(n/i);
		}
	}
}
ll Euler(ll n){//欧拉函数模板
    ll ans = 1;
    for(int i = 2; i * i <= n; i++){
        if(n % i == 0){
            n /= i;
            ans *= (i - 1);
            while(n % i == 0){
                n /= i;
                ans *= i;
            }
        }
    }
    if(n > 1) ans *= (n-1);
    return ans ;
}
//下面两个函数相当于快速幂取模，可以防止爆long long
ll Mul(ll x , ll y , ll mod){//快速幂取模模板
    ll ans = 0;
    while(y){
        if(y & 1){
            ans = (ans + x) % mod;
        }
        x = x * 2 % mod;
        y >>= 1;
    }
    return ans ;
}
ll pow_mod(ll x , ll n , ll mod ){
    ll ans = 1;
    ll t = x % mod ;
    while(n) {
        if(n & 1){
            ans = Mul(ans , t, mod);
        }
        t = Mul(t , t , mod);
        n >>= 1 ;
    }
    return ans ;
}
 
int main(){
    ll n;
    while(scanf("%lld",&n)!=EOF){
    	ve.clear();
        if(n % 2 == 0 || n == 1){
            printf("2^? mod %lld = 1\n",n);
            continue;
        }
       	ll ans = Euler(n);
       	get_prime(ans);
       	sort(ve.begin(),ve.end());
       	for(int i=0;i<ve.size();i++){
       		if( pow_mod(2ll,ve[i],n) == 1 ){
       			ans = ve[i];
       			break;
       		}
       	}
        printf("2^%lld mod %lld = 1\n", ans , n);
    }
    return 0;
}

HDU 2462 - 2^x mod n = 1（欧拉定理）【模板】

2^x mod n = 1

猜你喜欢