CERC2017 F-Faulty Factorial【数论】

题目描述

The factorial of a natural number is the product of all positive integers less than or equal to it. For example, the factorial of 4 is 1 · 2 · 3 · 4 = 24. A faulty factorial of length n is similar to the factorial of n, but it contains a fault: one of the integers is strictly smaller than what it should be (but still at least 1). For example, 1 · 2 · 2 · 4 = 16 is a faulty factorial of length 4. Given the length n, a prime modulus p and a target remainder r, find some faulty factorial of length n that gives the remainder r when divided by p.

Input

The first line contains three integers n, p and r (2 ≤ n ≤ 1018 , 2 ≤ p < 107 , 0 ≤ r < p) — the length of the faulty factorial, the prime modulus and the target remainder as described in the problem statement.

Output

If there is no faulty factorial satisfying the requirements output “-1 -1”. Otherwise, output two integers — the index k of the fault (2 ≤ k ≤ n) and the value v at that index (1 ≤ v < k). If there are multiple solutions, output any of them.

Sample Input 1

4 5 1

Sample Output 1

3 2

Sample Input 2

4 127 24

Sample Output 2

-1 -1

题意

题意还挺好理解的，给三个数，n,p,r（p > r）,在n的阶乘中找到一个数k（2<=k <= n）,将k换成比k小的数v（1<=v<k），使得换完之后的阶乘结果res对p取模结果为r，即res≡r(mod p)。

例如第一个样例，n = 4，阶乘为1*2*3*4，但若把3换成2，阶乘变为1*2*2*4 = 16 ≡ 1 (mod 5)

输出任意一组k，v，若找不到这样一组k，v，输出-1 -1

思路

这道题最好是讨论n，p的大小。

①n>=2p时，不管修改哪一个值，得到的结果总是p的倍数，r只能为0，所以r = 0时，修改任意一个值（例如使2改为1）即可，r ≠ 0，输出-1 -1

②p<=n<2p时，若r = 0，则修改的一定不是p，找到任意一个不是p的值改为1即可，找不到输出-1 -1（例如n = 2，p = 2，找不到）；若r ≠ 0，则修改的一定是p，至于修改成什么，可以枚举小于p的值，找到一个即可。

③n<p时，这时候用到数论的知识，根据res≡r(mod p)，可以整理成v≡(mod p),p是素数，根据费马小定理，求逆元

v ≡ ((r * k % p) * Fast Power(n!, p-2, p)) % p ,枚举k,若得到的v>=1&&v<k,则找到一组，若找不到输出-1 -1

代码

#include<iostream>
#include<cstdio>
using namespace std;

#define ll long long

ll fast_power(ll a, ll b, ll m)//快速幂
{
    ll ans = 1;
    a %= m;
    while(b)
    {
        if(b&1) ans = ans *a % m;
        a = a * a % m;
        b >>= 1;
    }
    return ans;
}
int main()
{
    ll n,p,r,v;
    bool flag = false;
    scanf("%lld %lld %lld", &n, &p, &r);
    if(n >= 2 * p)
    {
        if(r == 0)
            printf("2 1\n");
        else
            printf("-1 -1\n");
    }
    else if(n >= p && n < 2 * p)
    {
        if(r == 0)
        {
            for(ll i = 2; i <= n; i++)
            {
                if(i != p)//只要能找到不是p的k就可
                {
                    printf("%lld 1\n", i);
                    flag = true;
                    break;
                }
            }
            if(!flag)
                printf("-1 -1\n");
        }
        else
        {
            ll jc = 1;
            for(ll i = 2; i <= n; i++)
            {
                if(i == p)  continue;
                jc = (jc * (i % p)) % p;
            }
            for(ll i = 1; i < p; i++)//将k换成i
            {
                if((jc * (i % p)) % p == r)
                {
                    printf("%lld %lld\n", p, i);
                    flag = true;
                }
            }
            if(!flag)
                printf("-1 -1\n");
        }
    }
    else
    {
        ll jc = 1;
        for(ll i = 2; i <= n; i++)
            jc = (jc * (i % p)) % p;
        jc = fast_power(jc, p - 2, p);//求阶乘的逆元
        for(ll i = 2; i <= n; i++)
        {
            v = (((r % p) * (i % p)) % p * jc) % p;
            //printf("%lld\n", y);
            if(v >= 1 && v < i)
            {
                printf("%lld %lld\n", i, v);
                flag = true;
                break;
            }
        }
        if(!flag)
            printf("-1 -1\n");
    }
    //printf("%lld %lld %lld", n, p, r);
    return 0;
}

如有错误请指明~   ฅ●ω●ฅ