筛素数

#include <stdio.h>
#include <math.h>
struct point_node
{
    int x, y;
}point[1005];
int n;
bool func ( point_node a, point_node b, point_node c )
{
    if ( ( ( c.x - a.x ) * ( b.x - a.x ) + ( c.y - a.y ) * ( b.y - a.y ) == 0 ) &&
        ( ( c.x - a.x ) * ( c.x - a.x ) + ( c.y - a.y ) * ( c.y - a.y ) ==  ( b.x - a.x ) * ( b.x - a.x ) + ( b.y - a.y ) * ( b.y - a.y ) ) )
        return true;
    return false;
}
bool judge ( int a, int b )
{
    int flag[2] = {0};
    for ( int i = 0; i < n; i++ )
    {
        if ( i == a || i == b )
            continue;
        if ( func ( point[a], point[b], point[i] ) )
            flag[0]++;
        if ( func ( point[b], point[a], point[i] ) )
            flag[1]++;
    }
    if ( flag[0] &&  flag[1] )
        return true;
    return false;
}
int main ( )
{

    while ( scanf ( "%d", &n ), n )
    {
        int sum = 0, i, j;
        for ( i = 0; i < n; i++ )
            scanf ( "%d%d", &point[i].x, &point[i].y );
        if ( judge ( n - 1, 0 ) )
                    sum++;
        for ( i = 0; i < n; i++ )
            for ( j = i + 1; j < n; j++ )
                if ( judge ( i, j ) )
                    sum++;
        printf ( "%d\n", sum / 4 );
    }
    return 0;
}
430407199106240019<aaa_275@126.com>  21:50:56
#include <iostream>   
#include <ctime>   
#include <cmath>   
#define MAX_L 64               //最长位数   
#define TIMES 8                //miller robin素性测试的测试次数   
#define MAX_VAL (pow(2.0, 60)) //定义最大值   
#define CVAL 200   
using namespace std;   
  
//最小的素因子   
__int64 minFactor;   
  
//(1)计算a * b mod n, 思路: 利用b的二进制表示进行拆分计算   
//(2)例如: b = 1011101那么a * b mod n = (a * 1000000 mod n + a * 10000 mod n + a * 1000 mod n + a * 100 mod n + a * 1 mod n) mod n   
//(3)思路就是上面描述的那样, 那么可以用从低位往高位遍历b, 并用a来记录当前位为1的值，每次遇到b当前位为   
//1就将结果值加上a并 mod n，然后a 要乘以2   
__int64 multAndMod(__int64 a, __int64 b, __int64 n)   
{   
    a = a % n;   
    __int64 res = 0;   
  
    while(b)   
    {   
        //当前位为1   
        if(b & 1)   
        {   
            //加上当前权位值   
            res += a;   
            //相当于mod n   
            if(res >= n) res -= n;   
        }   
        //乘以2，提高一位   
        a = a<<1;   
        //mod n   
        if(a >= n) a -= n;   
        b = b >> 1;   
    }   
  
    return res;   
}   
  
//(1)计算a ^ b mod n, 思路: 和上面类似，也是利用b的二进制表示进行拆分计算   
//(2)例如: b = 1011101那么a ^ b mod n = [(a ^ 1000000 mod n) * (a ^ 10000 mod n) * (a ^ 1000 mod n) * (a ^ 100 mod n) * (a ^ 1 mod n)] mod n   
//(3)思路就是上面描述的那样, 那么可以用从低位往高位遍历b, 并用a来记录当前位为1的值，每次遇到b当前位为   
//1就将结果乘上a并 mod n，然后a 要乘以a以提升一位   
__int64 modAndExp(__int64 a, __int64 b, __int64 n)   
{   
    a = a % n;   
    __int64 res = 1;   
    while(b >= 1)   
    {   
        //遇到当前位为1，则让res * 当前a并mod n   
        if(b & 1)   
            res = multAndMod(res, a, n);   
        //a * a以提升一位   
        a = multAndMod(a, a, n);   
        b = b >> 1;   
    }   
    return res;   
}   
  
//MillerRobin素性测试，true:素数，flase:合数   
bool millerRobin(__int64 a, __int64 n)   
{   
    __int64 u = 0, cur = n - 1;   
    int t = 0;   
    bool find1 = false;   
    while(cur != 0)   
    {   
        if(!find1)   
        {   
            int pb = cur % 2;   
            if(pb == 0) t++;   
            else find1 = true;   
        }   
        if(find1)   
            break;   
        cur = cur / 2;   
    }   
    u = cur;   
  
    cur = modAndExp(a, u, n);   
    __int64 now;   
    for(int p = 1; p <= t; p++)   
    {   
        now = modAndExp(cur, 2, n);   
        if(cur != 1 && now == 1 && cur != n - 1)   
        {   
            //printf("%d %d\n", cur, now);   
            return false;   
        }   
        cur = now;   
    }   
    if(cur != 1)   
    {   
        //printf("a:%I64d u:%I64d n:%I64d val:%I64d\n", a, u, n, start);   
        return false;   
    }   
    //printf("a:%I64d u:%I64d n:%I64d val:%I64d\n", a, u, n, start);   
    return true;   
}   
  
//利用Miller Robin对n进行n次素性测试   
bool testPrime(int times, __int64 n)   
{   
    if(n == 2) return true;   
    if(n % 2 == 0) return false;   
  
    __int64 a; int t;   
    srand(time(NULL));   
    for(t = 1; t <= times; t++)   
    {   
        a = rand() % (n - 1) + 1;   
        if(!millerRobin(a, n)) return false;   
    }   
    return true;   
}   
  
__int64 gcd(__int64 a, __int64 b)   
{   
    if(b == 0) return (a);   
    return gcd(b, a % b);   
}   
  
__int64 PollardRho(__int64 n, int c)   
{   
    int i = 1;   
    srand(time(NULL));   
    __int64 x = rand() % n;   
    __int64 y = x;   
    int k = 2;   
    while(true)   
    {   
        i = i + 1;   
        x = (modAndExp(x, 2, n) + c) % n;   
        __int64 d = gcd(y - x, n);   
        if(1 < d && d < n) return d;   
        if(y == x) return n; //重复了, 说明当前x下无解，需要重新启动PollardRho   
        if(i == k)   
        {   
            y = x;   
            k = k * 2;   
        }   
    }   
}   
  
void getSmallest(__int64 n, int c)   
{   
    if(n == 1) return;   
    //判断当前因子是否为素数   
    if(testPrime(TIMES, n))   
    {   
        if(n < minFactor) minFactor = n;   
        return;   
    }   
    __int64 val = n;   
    //循环，知道找到一个因子   
    while(val == n)   
        val = PollardRho(n, c--);   
    //二分   
    getSmallest(val, c);   
    getSmallest(n / val, c);   
}   
int main()   
{   
    int caseN;   
    __int64 n;   
    scanf("%d", &caseN);   
    while(caseN--)   
    {   
        scanf("%I64d", &n);   
        minFactor = MAX_VAL;   
        if(testPrime(TIMES, n)) printf("Prime\n");   
        else  
        {   
            getSmallest(n, CVAL);   
            printf("%I64d\n", minFactor);   
        }   
    }   
    return 0;   
}  

Description

Given a big integer number, you are required to find out whether it's a prime number.

Input

The first line contains the number of test cases T (1 <= T <= 20 ), then the following T lines each contains an integer number N (2 <= N < 2⁵⁴).

Output

For each test case, if N is a prime number, output a line containing the word "Prime", otherwise, output a line containing the smallest prime factor of N.

Sample Input

2
5
10

Sample Output

Prime
2