loj #143. 质数判定

\(\color{#0066ff}{ 题目描述 }\)

判定输入的数是不是质数。

\(\color{#0066ff}{输入格式}\)

若干行，一行一个数 \(x\)。

行数不超过 \(10^5\)。

\(\color{#0066ff}{输出格式}\)

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 \(N\)。

\(\color{#0066ff}{输入样例}\)

1
2
6
9
666623333

\(\color{#0066ff}{输出样例}\)

N
Y
N
N
Y

\(\color{#0066ff}{数据范围与提示}\)

\(1≤x≤10^{18}\)。

\(\color{#0066ff}{ 题解 }\)

就是个Miller Rabbin的板子题

详解

#include
#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" : "N\n");
    return 0;
}

转载于:https://www.cnblogs.com/olinr/p/10305930.html