P5137-polynomial【倍增】

正题

题目链接:https://www.luogu.com.cn/problem/P5137

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题目大意

\(T\)组数据给出\(n,a,b,p\)求

\[\left(\sum_{0=1}^na^ib^{n-i}\right)\%p \]

\(1\leq T\leq 10^5,1\leq n,a,b,p\leq 10^{18}\)

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解题思路

这个数据很大，考虑倍增求。

设为答案\(f(n)\)，那么有

\[f(n)=f(\frac{n}{2})(a^{\frac{n}{2}}+b^{\frac{n}{2}})-a^{\frac{n}{2}}b^{\frac{n}{2}} \]

\[f(n)=af(n-1)+b^{n} \]

倍增维护就好了

时间复杂度\(O(T\log n)\)

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code

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cctype>
#define ll long long
using namespace std;
ll n,a,b,p,T;
ll read(){
	ll x=0,f=1;char c=getchar();
	while(!isdigit(c)){if(c=='-')f=-f;c=getchar();}
	while(isdigit(c)){x=(x<<1)+(x<<3)+c-'0';c=getchar();}
	return x*f;
}
ll fac(ll a,ll b){
	ll c=(long double)a*b/p;
	long double ans=a*b-c*p;
	if(ans>=p)ans-=p;
	else if(ans<0)ans+=p;
	return ans;
}
signed main()
{
	T=read();
	while(T--){
		n=read();a=read();b=read();p=read();
		ll ans=1,k=0,B=1,A=1;
		for(ll i=62;i>=0;i--){
			ans=(fac(A,ans)+fac(B,ans))%p;
			ans=(ans+p-fac(A,B))%p;
			k*=2;B=fac(B,B);A=fac(A,A);
			if((n>>i)&1){
				k++;B=fac(B,b);A=fac(A,a);
				ans=(fac(ans,a)+B)%p;
			}
		}
		printf("%lld\n",ans);
	}
	return 0;
}