标签：each 一个 mat 个数 this show mil esc about
Description:
Goldbach‘s conjecture is one of the oldest and best-known unsolved problems in number theory and all of mathematics. It states:
Every even integer greater than 2 can be expressed as the sum of two primes.
The actual verification of the Goldbach conjecture shows that even numbers below at least 1e14 can be expressed as a sum of two prime numbers.
Many times, there are more than one way to represent even numbers as two prime numbers.
For example, 18=5+13=7+11, 64=3+61=5+59=11+53=17+47=23+41, etc.
Now this problem is asking you to divide a postive even integer n (2<n<2^63) into two prime numbers.
Although a certain scope of the problem has not been strictly proved the correctness of Goldbach‘s conjecture, we still hope that you can solve it.
If you find that an even number of Goldbach conjectures are not true, then this question will be wrong, but we would like to congratulate you on solving this math problem that has plagued humanity for hundreds of years.
Input:
The first line of input is a T means the number of the cases.
Next T lines, each line is a postive even integer n (2<n<2^63).
Output:
The output is also T lines, each line is two number we asked for.
T is about 100.
本题答案不唯一，符合要求的答案均正确
1 8
3 5
题目大意就是给你一个偶数，让你把偶数分成两个素数和，输出任意一组答案即可
打表发现可能的分解情况中，较小的那个素数非常小，最大不过在一万左右
所以现在问题就变成了如何快速的判断一个数是否为素数
所以要用“Miller-Rabin素数检测算法”，具体参加如下博客
https://blog.csdn.net/zengaming/article/details/51867240
要注意的是题目给的数非常大，即使用long long存稍微算一下加法也会炸
所以要用unsigned long long 运算，输出用 %llu
#include <cstdio> #include <cstdlib> #include<iostream> #define N 10000 using namespace std; typedef unsigned long long ll; ll ModMul(ll a,ll b,ll n)//快速积取模 a*b%n { ll ans=0; while(b) { if(b&1) ans=(ans+a)%n; a=(a+a)%n; b>>=1; } return ans; } ll ModExp(ll a,ll b,ll n)//快速幂取模 a^b%n { ll ans=1; while(b) { if(b&1) ans=ModMul(ans,a,n); a=ModMul(a,a,n); b>>=1; } return ans; } bool miller_rabin(ll n)//Miller-Rabin素数检测算法 { ll i,j,a,x,y,t,u,s=10; if(n==2) return true; if(n<2||!(n&1)) return false; for(t=0,u=n-1;!(u&1);t++,u>>=1);//n-1=u*2^t for(i=0;i<s;i++) { a=rand()%(n-1)+1; x=ModExp(a,u,n); for(j=0;j<t;j++) { y=ModMul(x,x,n); if(y==1&&x!=1&&x!=n-1) return false; x=y; } if(x!=1) return false; } return true; } int main() { ll n; ll t; scanf("%llu",&t); while(t--) { scanf("%llu",&n); for(ll i=1;i<N;i++) if(miller_rabin(i)&&miller_rabin(n-i)) { printf("%llu %llu\n",i,n-i); break; } } return 0; }
标签：each 一个 mat 个数 this show mil esc about
原文地址：https://www.cnblogs.com/tian-luo/p/8909209.html
