German Collegiate Programming Contest 2015： K. Upside down primes

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Last night, I must have dropped my alarm clock. When the alarm went off in the morning, it showed $51:80$instead of $08:15$. This made me realize that if you rotate a seven segment display like it is used in digital clocks by $180$ degrees, some numbers still are numbers after turning them upside down.

My favourite numbers are primes, of course. Your job is to check whether a number is a prime and still a prime when turned upside down.

Input Format

One line with the integer $N$ in question (1 \le N \le 10^{16})(1≤N≤1016). $N$ will not have leading zeros.

Output Format

Print one line of output containing "yes" if the number is a prime and still a prime if turned upside down, "no"otherwise.

样例输入1

151

样例输出1

yes

样例输入2

23

样例输出2

no

样例输入3

18115211

样例输出3

no

注意刚开始也要判断这个数是不是素数

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<cstring>
#include<queue> 
#include<stdlib.h>
#include<cmath>
using namespace std;  
typedef long long ll;
using namespace std;  
/*ll Quick_Mod(ll a, ll b, ll mod)
{    
    ll res = 1,term = a % mod;    
    while(b)    
    {    
        if(b & 1) res = (res * term) % mod;    
        term = (term * term) % mod;    
        b >>= 1;    
    }    
    return res;    
}    
bool Is_Prime(ll n)    
{    
   int i;    
   srand(time(0));  
   for(i = 0;i < 30;i++)   
       if(Quick_Mod(1+rand()%(n-1) ,n - 1,n) != 1)    
         break;    
   if(i == 30) return true;    
   return false;    
}*/
bool isPrime(long long num) {
    if (num == 2 || num == 3) {
        return true;
    }

    //如果不在6的倍数附近，肯定不是素数
    if (num % 6 != 1 && num % 6 != 5) {
        return false;
    }
    //对6倍数附近的数进行判断
    int fz=sqrt(num)+1;
    for (int i = 5; i <=fz ; i += 6) {
        if (num % i == 0 || num % (i + 2) == 0) {
            return false;
        }
    }
    return true;
}
int main(){
	char s[20];
	scanf("%s",s);
	long long sum=0;
	int len=strlen(s);
	long long sum1=0;
	for(int i=0;i<len;i++){
		sum1=sum1*10+s[i]-'0';
	}
	if(!isPrime(sum1)){
		printf("no\n");
		return 0;
	}
	for(int i=len-1;i>=0;i--){
		if(s[i]=='0'||s[i]=='2'||s[i]=='5'||s[i]=='8'||s[i]=='1')
			sum=sum*10+ s[i]-'0';
		else if(s[i]=='6') sum=sum*10+9;
		else if(s[i]=='9') sum=sum*10+6;
		else{
			printf("no\n");
			return 0;
		}
	} 
	//cout<<sum<<endl;
	if(sum==1){
		printf("no\n");
		return 0;
	}
	if(isPrime(sum)){
		printf("yes\n");
	}
	else printf("no\n");
	return 0;
}

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