沟槽的组合变形

给定数 $p,m$，$p$ 是质数，求

$$\sum _{i=0}^{p-1}(\genfrac{}{}{0ex}{}{2i}{i})m^{i}modp$$

多测，$T\le {10}^4$，$1\le m<p\le {10}^{14}$。

---

忽略 $p=2$ 的情况，对 ${\displaystyle (\genfrac{}{}{0ex}{}{2n}{n})}$ 变形：

$$\begin{array}{rl}(\genfrac{}{}{0ex}{}{2n}{n})& =2^n\cdot \frac{1\times 3\times 5\times \cdots \times (2n-1)}{n!}\\ & \equiv (-2{)}^n\cdot \frac{\prod _{i=1}^n(p+1-2i)}{n!}\\ & \equiv (-4{)}^n\cdot \frac{\prod _{i=0}^{n-1}(\frac{p-1}{2}-i)}{n!}\\ & \equiv (-4{)}^n\cdot (\genfrac{}{}{0ex}{}{\frac{p-1}{2}}{n})\end{array}$$

原式即

$$\sum _{i=0}^{p-1}(-4m{)}^i(\genfrac{}{}{0ex}{}{\frac{p-1}{2}}{i})$$

二项式定理

$$(1-4m{)}^{\frac{p-1}{2}}$$

时间复杂度 $O(T\log p)$。

---

#include<bits/stdc++.h>
#define ll long long
#define N 2000010
using namespace std;
ll read(){
	ll x=0,w=1;char ch=getchar();
	while(!isdigit(ch)){if(ch=='-')w=-1;ch=getchar();}
	while(isdigit(ch))x=(x<<3)+(x<<1)+(ch^48),ch=getchar();
	return x*w;
}
ll p,m;
ll mul(ll x,ll y){
	return (__int128)x*y%p;
}
ll qpow(ll k,ll b){
	ll ret=1;
	while(b){
		if(b&1)ret=mul(ret,k);
		k=mul(k,k),b>>=1;
	}
	return ret;
}
void solve(){
	p=read(),m=read();
	if(p==2ll)return puts("1"),void();
	ll t=((1ll-4ll*m)%p+p)%p;
	printf("%lld\n",qpow(t,(p-1ll)/2));
}
int main(){
	int T=read();
	while(T--)solve();
	
	return 0;
}