Problem 51 - Find the smallest prime which, by changing the same part of the number, can form eight different primes.

링크

By replacing the 1^st digit of *3, it turns out that six of the nine possible values: 13, 23, 43, 53, 73, and 83, are all prime.

By replacing the 3^rd and 4^th digits of 56**3 with the same digit, this 5-digit number is the first example having seven primes among the ten generated numbers, yielding the family: 56003, 56113, 56333, 56443, 56663, 56773, and 56993. Consequently 56003, being the first member of this family, is the smallest prime with this property.

Find the smallest prime which, by replacing part of the number (not necessarily adjacent digits) with the same digit, is part of an eight prime value family.

힘든 문제였다.

파이썬에 permutation이 값이 중복되는 문제가 있어서 C++로 코딩했다.

C++

#include <cstdlib> #include <cstdio> #include <vector> #include <algorithm> #include <cmath> using namespace std; bool isprime(int n) { if (n == 1) return false; if (n < 4) return true; if ((n & 1) == 0) return false; if (n < 9) return true; if (n % 3 == 0) return false; int r = (int)sqrt(n); int f = 5; while (f <= r) { if ((n % f) == 0 || (n % (f + 2)) == 0) return false; f += 6; } return true; } int main() { //int last = 0; for (int last = 1; last <= 9; last += 2) { int q = 1; int res = -1; while (q < 300) { //printf("%d\n", q); if ((q*10 + last) % 3) { char sq[100] = {0}; sprintf(sq, "%d", q); int len = strlen(sq); vector<int> pattern(3 + len, 1); pattern[0] = pattern[1] = pattern[2] = 0; do { int falseCnt = 1-pattern[0]; int firstp = -1; for (int k = 1-pattern[0]; k < 10; k++) { int n = 0; int sqind = 0; for (int m = 0; m < (int)pattern.size(); m++) { n *= 10; if (pattern[m] == 0) n += k; else n += sq[sqind++] - '0'; } n = n * 10 + last; if (isprime(n) == false) { falseCnt++; if (falseCnt > 2) break; } else if (firstp == -1) firstp = n; } if (falseCnt <= 2) { printf("%d %d %d\n", q, last, firstp); break; } } while (next_permutation(pattern.begin(), pattern.end())); } q++; }} return 0; }