首页 > 代码库 > POJ1811_Prime Test【Miller Rabin素数測试】【Pollar Rho整数分解】
POJ1811_Prime Test【Miller Rabin素数測试】【Pollar Rho整数分解】
Prime Test
Time Limit: 6000MS
Memory Limit: 65536K
Total Submissions: 29193
Accepted: 7392
Case Time Limit: 4000MS
DescriptionGiven 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.
Sample Input
2
5
10
Sample Output
Prime
2
Source
POJ Monthly
题目大意:T组数据,对于输入的N,若N为素数,输出"Prime",否则输出N的最小素因子
思路:由于N的规模为2^54所以普通的素性推断果断过不了。
要用Miller Rabin素数測试来做。
而若N不为素数,则须要对N进行素因子分解。由于N为大数,考虑用Pollar Rho整数分解来做。
#include<stdio.h> #include<stdlib.h> #include<time.h> #include<math.h> #define MAX_VAL (pow(2.0,60)) //miller_rabbin素性測试 //__int64 mod_mul(__int64 x,__int64 y,__int64 mo) //{ // __int64 t; // x %= mo; // for(t = 0; y; x = (x<<1)%mo,y>>=1) // if(y & 1) // t = (t+x) %mo; // // return t; //} __int64 mod_mul(__int64 x,__int64 y,__int64 mo) { __int64 t,T,a,b,c,d,e,f,g,h,v,ans; T = (__int64)(sqrt(double(mo)+0.5)); t = T*T - mo; a = x / T; b = x % T; c = y / T; d = y % T; e = a*c / T; f = a*c % T; v = ((a*d+b*c)%mo + e*t) % mo; g = v / T; h = v % T; ans = (((f+g)*t%mo + b*d)% mo + h*T)%mo; while(ans < 0) ans += mo; return ans; } __int64 mod_exp(__int64 num,__int64 t,__int64 mo) { __int64 ret = 1, temp = num % mo; for(; t; t >>=1,temp=mod_mul(temp,temp,mo)) if(t & 1) ret = mod_mul(ret,temp,mo); return ret; } bool miller_rabbin(__int64 n) { if(n == 2) return true; if(n < 2 || !(n&1)) return false; int t = 0; __int64 a,x,y,u = n-1; while((u & 1) == 0) { t++; u >>= 1; } for(int i = 0; i < 50; i++) { a = rand() % (n-1)+1; x = mod_exp(a,u,n); for(int j = 0; j < t; j++) { y = mod_mul(x,x,n); if(y == 1 && x != 1 && x != n-1) return false; x = y; } if(x != 1) return false; } return true; } //PollarRho大整数因子分解 __int64 minFactor; __int64 gcd(__int64 a,__int64 b) { if(b == 0) return a; return gcd(b, a % b); } __int64 PollarRho(__int64 n, int c) { int i = 1; srand(time(NULL)); __int64 x = rand() % n; __int64 y = x; int k = 2; while(true) { i++; x = (mod_exp(x,2,n) + c) % n; __int64 d = gcd(y-x,n); if(1 < d && d < n) return d; if(y == x) return n; if(i == k) { y = x; k *= 2; } } } void getSmallest(__int64 n, int c) { if(n == 1) return; if(miller_rabbin(n)) { if(n < minFactor) minFactor = n; return; } __int64 val = n; while(val == n) val = PollarRho(n,c--); getSmallest(val,c); getSmallest(n/val,c); } int main() { int T; __int64 n; scanf("%d",&T); while(T--) { scanf("%I64d",&n); minFactor = MAX_VAL; if(miller_rabbin(n)) printf("Prime\n"); else { getSmallest(n,200); printf("%I64d\n",minFactor); } } return 0; }
POJ1811_Prime Test【Miller Rabin素数測试】【Pollar Rho整数分解】
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