Hackerrank--Prime Sum

题目链接

The problem is quite simple. You're given a number N and a positive integer K. Tell if N can be represented as a sum of K prime numbers (not necessarily distinct).

Input Format
The first line contains a single integer T, denoting the number of test cases. 
Each of the next T lines contains two positive integers, N & K, separated by a single space.

Output Format
For every test case, output "Yes" or "No" (without quotes).

Constraints
1 <= T <= 5000 
1 <= N <= 1012 
1 <= K <= 1012

Sample Input

2 10 2 1 6 

Sample Output

Yes No 

Explanation

In the first case, 10 can be written as 5 + 5, and 5 is a prime number. In the second case, 1 cannot be represented as a sum of prime numbers, because there are no prime numbers less than 1.

题意:给两个正整数n和k,问能否将n分解为k个素数的和(可以出现相同的)。

思路:本题涉及的知识点有哥德巴赫猜想(任何大于2的偶数都可以拆分成两个素数的和),

还有Miller-Rabin素数测试,一般使用的素数测试是O(sqrt(n))复杂度的,无法满足大整数的要求。

 

费马小定理如果p是一个素数,(0<a<p),

 

 

 

例如,67是一个素数,2^66 mod 67=1.

 

利用费马小定理,对于给定的整数n,可以设计一个素数判定算法.通过计算d=2^(n-1) mod n 来判定整数n的素性.d1,n肯定不是素数;d=1,n则很可能是素数,但也存在合数n,使得 .例如,满足此条件的最小合数是n=341.为了提高测试的准确性,我们可以随机地选取整数1<a<n-1,然后用条件 来判定整数n的素性.例如对于n=341,a=3, ,故可判定n不是素数.

 

       费马小定理毕竟只是素数判定的一个必要条件.满足费马小定理条件的整数n未必全是素数.有些合数也满足费马小定理的条件.这些合数被称作Carmichael,3Carmichael数是561,1105,1729. Carmichael数是非常少的.1~100000000范围内的整数中,只有255Carmichael.

 

       利用下面的二次探测定理可以对上面的素数判定算法作进一步改进,以避免将Carmichael数当作素数.

 

二次探测定理  如果p是一个素数,且0<x<p,则方程x*x1(mod p)的解为x=1,p-1.

 

       事实上, x*x1(mod p)等价于 x*x-10(mod p).由此可知;

 

        (x-1)(x+1) 1(mod p)

 

p必须整除x-1或x+1.由p是素数且 0<x<p,推出x=1或x=p-1.

 

       利用二次探测定理,我们可以在利用费马小定理计算 a^(n-1) mod n的过程中增加对于整数n的二次探测.一旦发现违背二次探测条件,即可得出n不是素数的结论.

 

 

Accepted Code:

 1 #include <ctime>

 2 #include <iostream>

 3 using namespace std;  4 

 5 typedef long long LL;  6 

 7 LL mulMod(LL a, LL b, LL c) {  8     LL res = 0;  9     while (b) { 10         if (b&1) if ((res = (res + a)) >= c) res -= c; 11         a = a + a; 12         if (a >= c) a -= c; 13         b >>= 1; 14  } 15     return res; 16 } 17 

18 LL powMod(LL a, LL b, LL c) { 19     LL res = 1; 20     while (b) { 21         if (b&1) res = mulMod(res, a, c); 22         a = mulMod(a, a, c); 23         b >>= 1; 24  } 25     return res; 26 } 27 

28 bool isPrime(LL n) { 29     if (n <= 1) return false; 30     if (n == 2) return true; 31     if (n & 1 == 0) return false; 32     srand((LL)time(0)); 33     LL u = n - 1, k = 0, pre; 34     while (!(u&1)) u >>= 1, k++; 35     for (int t = 0; t < 10; t++) { 36         LL a = rand() % (n - 2) + 2; 37         LL ans = powMod(a, n - 1, n); 38         for (int i = 0; i < k; i++) { 39             pre = ans; 40             ans = mulMod(ans, ans, n); 41             if (ans == 1 && (pre != 1 && pre != n - 1)) return false; 42             pre = ans; 43  } 44         if (ans != 1) return false; 45  } 46     return true; 47 } 48 

49 int main(void) { 50     ios::sync_with_stdio(false); 51     int T; 52     cin >> T; 53     while (T--) { 54  LL n, k; 55         cin >> n >> k; 56         if (n < 2 * k) { 57             cout << "No" << endl; 58         } else { 59             if (k == 1) { 60                 if (isPrime(n)) cout << "Yes" << endl; 61                 else cout << "No" << endl; 62             } else if (k == 2) { 63                 if (n % 2 == 0) cout << "Yes" << endl; 64                 else if (isPrime(n - 2)) cout << "Yes" << endl; 65                 else cout << "No" << endl; 66             } else { 67                 cout << "Yes" << endl; 68  } 69  } 70  } 71     return 0; 72 }

 

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