hdu 2138(米勒—拉賓素數測試)
阿新 • • 發佈:2019-02-16
How many prime numbers
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Problem Description Give you a lot of positive integers, just to find out how many prime numbers there are.
Input There are a lot of cases. In each case, there is an integer N representing the number of integers to find. Each integer won’t exceed 32-bit signed integer, and each of them won’t be less than 2.
Output For each case, print the number of prime numbers you have found out.
Sample Input 3 2 3 4
Sample Output 2 解題思路:這道題目如果靠篩選法那肯定會超時。米勒—拉賓素數測試可以進行大素數判斷。。。 米勒—拉賓素數測試主要依靠兩個基本定理: 1)費馬小定理:當gcd(a,p)=1且p為素數時,有a^p-1 mod p = 1 2)快速冪取模演算法(RSA公鑰加密演算法) 其實這個演算法模板我也沒有很明白,先湊合著看吧。。。 AC:
#include<iostream> #include<cstdio> #include<cstring> using namespace std; __int64 qpow(int a,int b,int r) { __int64 ans = 1,buff = a; while(b) { if(b & 1) ans = (ans*buff) % r; buff = (buff*buff) % r; b >>= 1; } return ans; } bool Miller_Rabbin(int n,int a) { int r = 0,s = n-1,j; if(!(n % a)) return false; while(!(s & 1)) { s >>= 1; r++; } __int64 k = qpow(a,s,n); if(k == 1) return true; for(j = 0; j < r; j++, k = k * k % n) if(k == n-1) return true; return false; } bool isPrime(int n) { int tab[4] = {2,3,5,7}; for(int i = 0; i < 4; i++) { if(n == tab[i]) return true; if(!Miller_Rabbin(n,tab[i])) return false; } return true; } int main() { int n; while(cin>>n) { int a,ans = 0; while(n--) { cin>>a; if(isPrime(a)) ans++; } cout<<ans<<endl; } return 0; }