LOJ#162. 快速幂 2（分块）

题面

传送门

题解

orzljz

我们分块，设\(s=\sqrt{p}+1\)，那么\(x^a\)可以拆成\((x^s)^{a/s}\)和\(x^{a\bmod s}\)，\(O(s)\)预处理，\(O(1)\)计算就可以了

//minamoto
#include<bits/stdc++.h>
#define R register
#define inline __attribute__((always_inline))
#define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)
#define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
using namespace std;
char buf[1<<21],*p1=buf,*p2=buf;
inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}
int read(){
    R int res,f=1;R char ch;
    while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);
    for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');
    return res*f;
}
char sr[1<<21],z[20];int C=-1,Z=0;
inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}
void print(R int x){
    if(C>1<<20)Ot();if(x<0)sr[++C]='-',x=-x;
    while(z[++Z]=x%10+48,x/=10);
    while(sr[++C]=z[Z],--Z);sr[++C]=' ';
}
const int N=50005,P=998244352;
inline int add(R int x,R int y){return x+y>=P?x+y-P:x+y;}
inline int dec(R int x,R int y){return x-y<0?x-y+P:x-y;}
inline int mul(R int x,R int y){return 1ll*x*y-1ll*x*y/P*P;}
int ksm(R int x,R int y){
	R int res=1;
	for(;y;y>>=1,x=mul(x,x))(y&1)?res=mul(res,x):0;
	return res;
}
int bin[N],bs[N],n,x,s,a;
int main(){
//	freopen("testdata.in","r",stdin);
	x=read(),n=read(),s=sqrt(P)+1;
	bin[0]=1;fp(i,1,s)bin[i]=mul(bin[i-1],x);
	bs[0]=1;fp(i,1,s)bs[i]=mul(bs[i-1],bin[s]);
	while(n--)a=read(),print(mul(bs[a/s],bin[a%s]));
	return Ot(),0;
}