2^x mod n = 1

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20711    Accepted Submission(s): 6500

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

2
5

Sample Output

2^? mod 2 = 1
2^4 mod 5 = 1

Author

MA, Xiao

Source

ZOJ Monthly, February 2003

Recommend

Ignatius.L

能不能不用暴力

好的 请转到https://www.cnblogs.com/WTSRUVF/p/10798178.html

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <list>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define rb(a) scanf("%lf", &a)
#define rf(a) scanf("%f", &a)
#define pd(a) printf("%d\n", a)
#define plld(a) printf("%lld\n", a)
#define pc(a) printf("%c\n", a)
#define ps(a) printf("%s\n", a)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 1100000, INF = 0x7fffffff;
int prime[maxn + 10], phi[maxn + 10];
bool vis[maxn + 10];
int n, ans;

void getphi()
{
    ans = 0;
    phi[1] = 1;
    for(int i = 2; i <= maxn; i++)
    {
        if(!vis[i])
        {
            prime[++ans] = i;
            phi[i] = i - 1;
        }
        for(int j = 1; j <= ans; j++)
        {
            if(i * prime[j] > maxn) break;
            vis[i * prime[j]] = 1;
            if(i % prime[j] == 0)
            {
                phi[i * prime[j]] = phi[i] * prime[j];
                break;
            }
            else
                phi[i * prime[j]] = phi[i] * (prime[j] - 1);

        }
    }
}

int q_pow(int a, int b, int mod)
{
    int res = 1;
    while(b)
    {
        if(b & 1) res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}

vector<int> v;

int main()
{
    getphi();
    while(cin >> n)
    {
        v.clear();
        if(n % 2 == 0 || n == 1)
        {
            printf("2^? mod %d = 1\n", n);
            continue;
        }
        int res = phi[n];
        for(int i = 2; i <= sqrt(phi[n] + 0.5); i++)
        {
            if(res % i == 0)
                v.push_back(i), v.push_back(res / i);
        }
        sort(v.begin(), v.end());
        for(int i = 0; i < v.size(); i++)
        {
            if(q_pow(2, v[i], n) == 1)
            {
                res = v[i];
                break;
            }

        }

                printf("2^%d mod %d = 1\n", res, n);

    }

    return 0;

}

2^x mod n = 1

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20711    Accepted Submission(s): 6500

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

2
5

Sample Output

2^? mod 2 = 1
2^4 mod 5 = 1

Author

MA, Xiao

Source

ZOJ Monthly, February 2003

Recommend

Ignatius.L

原文地址：https://www.cnblogs.com/WTSRUVF/p/10800970.html