【JZOJ】【数论】Idiot 的乘幂

$Link$Link

$JZOJ$JZOJ $3909$3909

$Description$Description

$Hint$Hint

$Train$Train $of$of $Thought$Thought

将题目中的方程转化为
$X ≡ X^{ax+cy} ≡ b^xd^y$X≡Xax+cy≡bxdy
然后通过一次扩欧（扩展欧几里得）求出可能的$x$x和$y$y
算出$X$X再套回原方程检验就好了
但是，又因为求出来的$x$x和$y$y可能是负数
这种情况就还要给$b$b和$d$d取一个逆元（求法还是扩欧）

$Code$Code

#include<iostream>
#include<cstdio>
#define ll long long
 
using namespace std;

int t;
ll a, b, c, d, p;

ll exgcd(ll a, ll b, ll &x, ll &y)
{
	if (b == 0) {
		x = 1; y = 0;
		return a;
	}
	int val = exgcd(b, a % b, x, y);
	ll z = x;
	x = y, y = z - y * (a / b);
	return val; 
}//扩欧

ll qpower(ll a, ll b)
{
	ll ans = 1;
	a %= p;
	for (; b > 0; b >>= 1)
	{
		if (b & 1) ans = (ll) (ans * a + p) % p;
		a = (ll) (a * a + p) % p;
	}
	return ans;
}

int main()
{
	scanf("%d", &t);
	for (int i = 1; i <= t; ++i)
	{
		ll x = 0, y = 0;
		scanf("%lld%lld%lld%lld%lld", &a, &b, &c, &d, &p);
		exgcd(a, c, x, y);
		
		ll bf = b, df = d, unknown;
		if (x < 0) {
			exgcd(b, p, bf, unknown); //求b的逆元
			x = -x; 
		} 
		if (y < 0) {
			exgcd(d, p, df, unknown);//求d的逆元
			y = -y; 
		}
		
		ll X = (qpower(bf % p, x) % p * qpower(df % p, y) % p) % p;
		if ((qpower(X, a) % p == b % p) && (qpower(X, c) % p == d % p))//原方程检验
		 printf("%lld\n", X);
		 else printf("No Solution!\n");
	}
}