PAT 甲级 1015 Reversible Primes (20分)

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最新推荐文章于 2022-07-27 16:13:26 发布

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题目

1015 Reversible Primes (20分)

A reversible prime in any number system is a prime whose “reverse” in that number system is also a prime. For example in the decimal system 73 is a reversible prime because its reverse 37 is also a prime.

Now given any two positive integers N (<105) and D (1<D≤10), you are supposed to tell if N is a reversible prime with radix D.

Input Specification:

The input file consists of several test cases. Each case occupies a line which contains two integers N and D. The input is finished by a negative N.

Output Specification:

For each test case, print in one line Yes if $N$ is a reversible prime with radix $D$ , or No if not.

Sample Input:

Sample Output:

Yes
Yes
No

题目大意

给出一个数N和一个表示进制的数D，如果该数和在对应进制下反转后的数字都是素数，则输出Yes，否则No

思路

首先可以通过欧拉筛法来建立一个素数判断表；对于给定的数和表示进制的数，获得其在对应进制数下的表示，可以存放在字符串中，得到反转后的数并判断是否是素数。

代码中字符串s获得的时候已经是反转的情况

代码

#include<bits/stdc++.h>
using namespace std;
const int maxN = 1000010;
int ifsu[maxN]={0};
void initial(){
    for(int i=2; i<maxN; i++){
        if(!ifsu[i]){ // 是素数
            for(int j=i+i; j<maxN; j+=i){
                ifsu[j] = 1;
            }
        }
    }
}
int getNum(int a, int radix){
    string s = "";
    while(a != 0){
        s += (a % radix + '0');
        a = a/radix;
    }
    int sum = 0, p=0;
    for(int i=s.size()-1; i>=0; i--){
        sum += pow(radix, p++) * (s[i]-'0');
    }
    return sum;
}
int main(int argc, const char * argv[]) {
    int a, radix;
    ifsu[1] = 1;
    initial();
    while(true){
        scanf("%d", &a);
        if(a < 0) break;
        scanf("%d", &radix);
        if(!ifsu[a]){
            int sum = getNum(a, radix);
            if(!ifsu[sum]) printf("Yes\n");
            else printf("No\n");
        }else
            printf("No\n");
    }
    return 0;
}