POJ1811(SummerTrainingDay04-G miller-rabin判断素性 && pollard-rho分解质因数)
Prime Test
| Time Limit: 6000MS | Memory Limit: 65536K | |
| Total Submissions: 35528 | Accepted: 9479 | |
| Case Time Limit: 4000MS | ||
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
Source
1 //2017-08-16 2 #include <cstdio> 3 #include <cstring> 4 #include <iostream> 5 #include <algorithm> 6 #define ll long long 7 8 using namespace std; 9 10 const int TIMES = 10; 11 12 ll random0(ll n){ 13 return ((double)rand() / RAND_MAX*n + 0.5); 14 } 15 16 //快速乘a*b%mod 17 ll quick_mul(ll a, ll b, ll mod){ 18 ll ans = 0; 19 while(b){ 20 if(b&1){ 21 b--; 22 ans = (ans+a)%mod; 23 } 24 b >>= 1; 25 a = (a+a) % mod; 26 } 27 return ans; 28 } 29 30 //快速幂a^b%mod 31 ll quick_pow(ll a, ll n, ll mod){ 32 ll ans = 1; 33 while(n){ 34 if(n&1)ans = quick_mul(ans, a, mod); 35 a = quick_mul(a, a, mod); 36 n >>= 1; 37 } 38 return ans; 39 } 40 41 bool witness(ll a, ll n){ 42 ll tmp = n-1; 43 int i = 0; 44 while(tmp % 2 == 0){ 45 tmp >>= 1; 46 i++; 47 } 48 ll x = quick_pow(a, tmp, n); 49 if(x == 1 || x == n-1)return true; 50 while(i--){ 51 x = quick_mul(x, x, n); 52 if(x == n-1)return true; 53 } 54 return false; 55 } 56 57 bool miller_rabin(ll n){ 58 if(n == 2)return true; 59 if(n < 2 || n % 2 == 0)return false; 60 for(int i = 1; i <= TIMES; i++){ 61 ll a = random0(n-2)+1; 62 if(!witness(a, n)) 63 return false; 64 } 65 return true; 66 } 67 68 //factor存放分解出来的素因数 69 ll factor[1000]; 70 int tot; 71 72 ll gcd(ll a, ll b){ 73 if(a == 0)return 1; 74 if(a < 0)return gcd(-a, b); 75 while(b){ 76 ll tmp = a % b; 77 a = b; 78 b = tmp; 79 } 80 return a; 81 } 82 83 ll pollard_rho(ll x, ll c){ 84 ll i = 1, k = 2; 85 ll x0 = rand()%x, x1 = x0; 86 while(1){ 87 i++; 88 x0 = (quick_mul(x0, x0, x)+c)%x; 89 ll d = gcd(x1-x0, x); 90 if(d != 1 && d != x)return d; 91 if(x1 == x0)return x; 92 if(i == k){ 93 x1 = x0; 94 k += k; 95 } 96 } 97 } 98 99 //对n分解质因数 100 void decomposition_factor(ll n){ 101 if(miller_rabin(n)){ 102 factor[tot++] = n; 103 return; 104 } 105 ll p = n; 106 while(p >= n){ 107 p = pollard_rho(p, rand()%(n-1)+1); 108 } 109 decomposition_factor(p); 110 decomposition_factor(n/p); 111 } 112 113 int main() 114 { 115 int T; 116 scanf("%d", &T); 117 while(T--){ 118 ll n; 119 scanf("%lld", &n); 120 if(miller_rabin(n)) 121 printf("Prime\n"); 122 else{ 123 tot = 0; 124 decomposition_factor(n); 125 ll ans = factor[0]; 126 for(int i = 1; i < tot; i++) 127 if(factor[i] < ans)ans = factor[i]; 128 printf("%lld\n", ans); 129 } 130 } 131 132 return 0; 133 }
