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pandigital_prime.cpp
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pandigital_prime.cpp
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#include <iostream>
using namespace std;
int length(long long int a) {
int count = 0;
while(a > 0) {
a /= 10;
count++;
}
return count;
}
bool is_pandigital(long long int a) {
int l = length(a);
long long int b; int chk;
while(l > 0) {
chk = 1;
b = a;
while(b > 0) {
if(l == b % 10) {
chk = 0;
break;
}
b /= 10;
}
if(chk) return false;
l--;
}
return true;
}
bool is_prime(long long int a) {
for(int i = 2; i * i <= a; i++) if(a % i == 0) return false;
return true;
}
int main(void) {
for(long long int i = 7654321; i >= 0; i--) {
if(is_pandigital(i)) if(is_prime(i)) {
cout << i << endl;
break;
}
else continue;
}
return 0;
}
/*
NOTES: The insight in this problem is pandigit numbers can only be of less than 10 digits, as to make a 10 digit number you must use zero, but using zero defys the very definition of pandigit. All pandigit numbers of 9 digits can't be primes as the sum of the digits is (9 * 10) / 2, which is divisible by 3.
All 9 digit pandigit numbers are divisible by 3, as the sum of the digits will be 45.
All 8 digit pandigit numbers are divisible by 3, as the sum of the digits will be 36, which again is divisible by 3.
All 6 digit pandigit numbers are also divisible by 3, as the sum of the digits will be 21.
So, the highest pandigit number which is also prime must be a 7 digit number
The strategy here is to start checking from the highest 7 digit pandigit number, and halt till you obtain the prime.
*/