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bzoj 3667

3667: Rabin-Miller算法

Time Limit: 60 Sec  Memory Limit: 512 MB
Submit: 1214  Solved: 368
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Description

 

Input

第一行:CAS,代表数据组数(不大于350),以下CAS行,每行一个数字,保证在64位长整形范围内,并且没有负数。你需要对于每个数字:第一,检验是否是质数,是质数就输出Prime 
第二,如果不是质数,输出它最大的质因子是哪个。 

Output

第一行CAS(CAS<=350,代表测试数据的组数) 
以下CAS行:每行一个数字,保证是在64位长整形范围内的正数。 
对于每组测试数据:输出Prime,代表它是质数,或者输出它最大的质因子,代表它是和数 

Sample Input

6
2
13
134
8897
1234567654321
1000000000000

Sample Output

Prime
Prime
67
41
4649
5

HINT

 

数据范围: 

保证cas<=350,保证所有数字均在64位长整形范围内。 

 裸的rho。跑得好慢,30多秒,是不是我乘法那一块比别人慢啊?

 

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<algorithm>
#define rep(i,j,k) for(register int i = j; i <= k; i++)
#define dow(i,j,k) for(register int i = j; i >= k; i--)
#define ll long long
using namespace std;
 
inline ll read() {
    ll s = 0, t = 1; char c = getchar();
    while( !isdigit(c) ) { if( c == ‘-‘ ) t = -1; c = getchar(); }
    while( isdigit(c) ) s = s * 10 + c - 48, c = getchar();
    return s * t; 
}
 
const int times = 15;
inline void M(ll &x,ll p) { if( x >= p ) x -= p; }
inline void Max(ll &x,ll v) { if( v > x ) x = v; }
inline ll mul(ll x,ll t,ll p) {
    ll ret = 0;
    while( t ) {
        if( t & 1 ) M(ret += x,p);
        M(x += x,p), t >>= 1; 
    } return ret;
}
inline ll pow(ll x,ll t,ll p) {
    ll ret = 1;
    while( t ) {
        if( t & 1 ) ret = mul(ret,x,p);
        x = mul(x,x,p), t >>= 1;
    } return ret;
}
inline bool miller_rabin(ll n) {
    if( n == 2 || n == 3 || n == 5 ) return 1;
    if( n < 2 || !(n&1) || n % 3 == 0 || n % 5 == 0 ) return 0;
    ll m = n - 1; int k = 0;
    while( !(m & 1) ) m >>= 1, k++;
    rep(i,1,times) {
        ll x = pow(rand() % (n-1) + 1,m,n), y;
        rep(j,1,k) {
            y = mul(x,x,n);
            if( y == 1 && x != 1 && x != n - 1 ) return 0;
            x = y; 
        }
        if( y != 1 ) return 0;
    }
    return 1;
}
 
inline ll gcd(ll x,ll y) {
    ll t;
    while( y ) t = x, x = y, y = t % y;
    return x;
}
inline ll rho(ll n,int c) {
    ll i = 1, k = 2, x = rand() % (n-1) + 1, y = x;
    while( 1 ) {
        i++;
        x = (mul(x,x,n) + c) % n;
        ll d = gcd((y-x+n)%n,n);
        if( 1 < d && d < n ) return d;
        if( x == y ) return n;
        if( i == k ) y = x, k <<= 1;
    }
}
 
ll ans = 0;
inline void find(ll x,int k) {
    if( x == 1 || x <= ans ) return;
    if( miller_rabin(x) ) { Max(ans,x); return; }
    ll p = x; int c = k;
    while( p >= x ) p = rho(x,k--);
    find(p,c), find(x/p,c);
} 
 
int main() {
    int T = read();
    while( T-- ) {
        ll n = read(); ans = 0;
        find(n,120);
        if( ans == n ) puts("Prime");
        else printf("%lld\n",ans);
    }
    return 0;
}

  

bzoj 3667