PTA (Advanced Level) 1015 Reversible Primes
Reversible Primes
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 (<) and D (1), 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:
73 10
23 2
23 10
-2
Sample Output:
Yes
Yes
No
解题思路:
本题给出一个十进制数字n,之后给出其进制d,要求判断其是不是可逆素数,若是可逆素数输出Yes否则输出No。
本题的可逆素数要求给出数字本身是素数且,将其转化为给出的进制,反转后重新转化为十进制还是素数。
如:23的2进制为10111反转后为11101 = 29,29与23都是素数,所以23是可逆素数。
由于本题有多组输入,所以可以用素数筛(埃氏筛法)先将素数打表。
埃氏筛法:
int prime[maxn];
bool vis[maxn] = {false};
int cnt = ;
void findPrime(){ //埃氏筛法
//每找到一个一个素数将其倍数都标记为不是素数
//时间复杂度O(n loglogn)
for(int i = ; i < maxn; i++){
if(vis[i] == false){ //i是素数
prime[cnt++] = i;
for(int j = * i; j < maxn; j += i){
vis[j] = true; //标记所以i的倍数
}
}
}
}
之后根据我们得到的素数判断输入的数是否为素数,若不是素数直接输出No,若是素数则将其转化为对应进制的数字后反转,将反转后得到的数字重新转化为10进制,在判断其是不是素数,是的话输出Yes否则输出No。
这样我们就需要两个函数,一个用来将其他进制数转化为10进制。另一个用来将10进制转化为其他进制并反转。
转化为d进制:
string decimalToOther(int num, int radix){ //将十进制数转化为其他进制数
string ans;
//ans从最低位开始记录,结束后得到的直接就是转化后数字的反转
while(num){
ans += (num % radix) + '';
num /= radix;
}
return ans;
}
转化为10进制
LL toDecimal(string num, int radix){ //将某进制数转化为10进制
LL ans = ;
for(int i = ; i < num.size(); i++){
ans = ans * radix + (num[i] - '');
if(ans < )
return -;
}
return ans;
}
AC代码
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 1e6+;
int prime[maxn];
bool vis[maxn] = {false};
int cnt = ;
void findPrime(){ //埃氏筛法
//每找到一个一个素数将其倍数都标记为不是素数
//时间复杂度O(n loglogn)
for(int i = ; i < maxn; i++){
if(vis[i] == false){ //i是素数
prime[cnt++] = i;
for(int j = * i; j < maxn; j += i){
vis[j] = true; //标记所以i的倍数
}
}
}
}
LL toDecimal(string num, int radix){ //将某进制数转化为10进制
LL ans = ;
for(int i = ; i < num.size(); i++){
ans = ans * radix + (num[i] - '');
if(ans < )
return -;
}
return ans;
}
string decimalToOther(int num, int radix){ //将十进制数转化为其他进制数
string ans;
//ans从最低位开始记录,结束后得到的直接就是转化后数字的反转
while(num){
ans += (num % radix) + '';
num /= radix;
}
return ans;
}
int n, d;
int main()
{
findPrime(); //素数打表
while(scanf("%d", &n) != EOF && n > ){
//输入n,n < 0时直接结束运算
scanf("%d", &d);
//输入进制d
if(vis[n] == true || n <= ){ //若n不是素数直接输出No进行下一次运算
printf("No\n");
continue;
}
string toRadix = decimalToOther(n, d);
//将n转化为d进制并反转
int ans = toDecimal(toRadix, d);
//将反转的数重新转化为10进制
if(vis[ans] == true || ans <= ){//若反转的数不是素数输出No
printf("No\n");
continue;
}
printf("Yes\n"); //否则输出Yes
}
return ;
}
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