POJ 1811 Prime Test（Miller-Rabin & Pollard-rho素数测试）

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 < 2 ⁵⁴).

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.

 

题目大意：给一个数n，问是不是素数，若是输出Prime，若不是输出其最小的非1因子。

思路：http://www.2cto.com/kf/201310/249381.html

模板盗自：http://vfleaking.blog.163.com/blog/static/1748076342013231104455989/

 

代码（375MS）：

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <algorithm>

 4 #include <cstring>

 5 #include <iterator>

 6 #include <vector>

 7 using namespace std;

 8 

 9 typedef long long LL;

10 

11 LL modplus(LL a, LL b, LL mod) {

12    LL res = a + b;

13    return res < mod ? res : res - mod;

14 }

15 

16 LL modminus(LL a, LL b, LL mod) {

17    LL res = a - b;

18    return res >= 0 ? res : res + mod;

19 }

20 

21 LL mult(LL x, LL p, LL mod) {

22     LL res = 0;

23     while(p) {

24         if(p & 1) res = modplus(res, x, mod);

25         x = modplus(x, x, mod);

26         p >>= 1;

27     }

28     return res;

29 }

30 

31 LL power(LL x, LL p, LL mod) {

32     LL res = 1;

33     while(p) {

34         if(p & 1) res = mult(res, x, mod);

35         x = mult(x, x, mod);

36         p >>= 1;

37     }

38     return res;

39 }

40 

41 bool witness(LL n, LL p) {

42    int t = __builtin_ctz(n - 1);

43    LL x = power(p % n, (n - 1) >> t, n), last;

44    while(t--) {

45        last = x, x = mult(x, x, n);

46        if(x == 1 && last != 1 && last != n - 1) return false;

47    }

48    return x == 1;

49 }

50 

51 const int prime_n = 5;

52 int prime[prime_n] = {2, 3, 7, 61, 24251};

53 

54 bool isPrime(LL n) {

55    if(n == 1) return false;

56    if(find(prime, prime + prime_n, n) != prime + prime_n) return true;

57    if(n % 2 == 0) return false;

58    for(int i = 0; i < prime_n; i++)

59        if(!witness(n, prime[i])) return false;

60    return true;

61 }

62 

63 LL getDivisor(LL n) {

64    int c = 1;

65    while (true) {

66        int i = 1, k = 2;

67        LL x1 = 1, x2 = 1;

68        while(true) {

69            x1 = modplus(mult(x1, x1, n), c, n);

70            LL d = __gcd(modminus(x1, x2, n), n);

71            if(d != 1 && d != n) return d;

72            if(x1 == x2) break;

73            i++;

74            if(i == k) x2 = x1, k <<= 1;

75        }

76        c++;

77    }

78 }

79 

80 void getFactor(LL n, vector<LL> &ans) {

81     if(isPrime(n)) return ans.push_back(n);

82     LL d = getDivisor(n);

83     getFactor(d, ans);

84     getFactor(n / d, ans);

85 }

86 

87 int main() {

88     int T; LL n;

89     scanf("%d", &T);

90     while(scanf("%I64d", &n) != EOF) {

91         if(isPrime(n)) puts("Prime");

92         else {

93             vector<LL> ans;

94             getFactor(n, ans);

95             printf("%I64d\n", *min_element(ans.begin(), ans.end()));

96         }

97     }

98 }

View Code