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
$$\left(\sum _{0=1}^n{a}^i{b}^{n-i}\right)\%p$$

$1\le T\le 10^5,1\le n,a,b,p\le 10^{18}$

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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;
}