[Comet OJ - Contest # 13]. D leather fire mouse - no anxiety of heart -

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Description

Portal

$T <= 1e5 sets of data, each read n, a, b, p (1≤n, a, b, p≤10 ^ {18})$

Solution

• will $a^i$ into $(\sqrt{a})^{2i}$ , and i can be into 2i, then according to the form of this equation is clearly $(a+b)^n$ is the same
• But we are due to enumerate $2i$ , so we only need an even number of items. But we can see that if the number is odd, it must also leave $\sqrt{a}$。
• You only need to take $(\sqrt{a}+b)^n$ the last integer like, leaving a $k \sqrt{a}$The contribution of odd items.
• Fast power can be.

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define ll __int128
using namespace std;

int T;
ll A,B,n,p;

void read(ll &x){
	x=0; char ch=getchar();
	for(;ch<'0'||ch>'9';ch=getchar());
	for(;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0';
}

int pd[50];
void write(ll x){
	if (!x) {printf("0\n");return;}
	while (x) pd[++pd[0]]=x%10,x/=10;
	while (pd[0]) putchar(pd[pd[0]--]+'0');
	puts("");
}

struct num{
	ll x,y;
	num(ll _x,ll _y){x=_x,y=_y;}
};
num operator *(num a,num b){return num((a.x*b.y+b.x*a.y)%p,(a.x*b.x%p*A+a.y*b.y)%p);}

ll qp(){
	num s=num(0,1),x=num(1,B);
	for(;n;n/=2,x=x*x) if (n&1)
		s=s*x;
	return s.y;
}

int main(){
	freopen("ceshi.in","r",stdin);
	scanf("%d",&T);
	while (T--){
		read(n),read(A),read(B),read(p);
		A=A%p,B=B%p;
		write(qp());
	}
}