loj #143. 质数判定

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判定输入的数是不是质数。

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若干行，一行一个数 (x)。

行数不超过 (10^5)。

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对于输入的每一行，如果 (x) 是质数输出一行 (Y)，否则输出一行 (N)。

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1
2
6
9
666623333

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N
Y
N
N
Y

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(1≤x≤10^{18})。

(color{#0066ff}{ 题解 })

就是个Miller Rabbin的板子题

详解

#include<bits/stdc++.h>
#define LL long long
LL in() {
	char ch; LL x = 0, f = 1;
	while(!isdigit(ch = getchar()))(ch == '-') && (f = -f);
	for(x = ch ^ 48; isdigit(ch = getchar()); x = (x << 1) + (x << 3) + (ch ^ 48));
	return x * f;
}
int prime[] = {2, 3, 5, 7, 11, 61, 24251};
LL msc(LL x, LL y, LL mod) {
	x %= mod;
	y %= mod;
	LL c = (long double)x / mod * y;
	LL d = x * y - c * mod;
	return ((d % mod) + mod) % mod;
}
LL ksm(LL x, LL y, LL mod) {
	LL re = 1LL;
	while(y) {
		if(y & 1) re = msc(re, x, mod);
		x = msc(x, x, mod); 
		y >>= 1;
	}
	return (re + mod) % mod;
}
bool judge(LL a, LL p) {
	LL s = p - 1;
	while(!(s & 1)) s >>= 1;
	LL k = ksm(a, s, p);
	while (s != p - 1 && k != 1 && k != p - 1) k = msc(k, k, p), s <<= 1;
	return (k == p - 1) || ((s & 1));
}
bool judge(LL n) {
	if(n == 1) return false;
	for(int i = 0; i < 7; i++) {
		if(n == prime[i]) return true;
		if(n % prime[i] == 0) return false;
		if(!judge(prime[i], n)) return false;
	}
	for(int i = 1; i <= 10; i++) if(!judge(2 + rand() % (n - 2), n)) return false;
	return true;
}


int main() {
	LL n;
	while(~scanf("%lld", &n)) printf(judge(n)? "Y
" : "N
");
	return 0;
}