一. 原题

Superprime Rib

Butchering Farmer John's cows always yields the best prime rib. You can tell prime ribs by looking at the digits lovingly stamped across them, one by one, by FJ and the USDA. Farmer John ensures that a purchaser of his prime ribs gets really prime ribs because when sliced from the right, the numbers on the ribs continue to stay prime right down to the last rib, e.g.:

7 3 3 1

The set of ribs denoted by 7331 is prime; the three ribs 733 are prime; the two ribs 73 are prime, and, of course, the last rib, 7, is prime. The number 7331 is called a superprime of length 4.

Write a program that accepts a number N 1 <=N<=8 of ribs and prints all the superprimes of that length.

The number 1 (by itself) is not a prime number.

PROGRAM NAME: sprime

INPUT FORMAT

A single line with the number N.

SAMPLE INPUT (file sprime.in)

4

OUTPUT FORMAT

The superprime ribs of length N, printed in ascending order one per line.

SAMPLE OUTPUT (file sprime.out)

2333 2339 2393 2399 2939 3119 3137 3733 3739 3793 3797 5939 7193 7331 7333 7393

二. 分析

根据每一位可能的取值构造n位数，最高位只能是{2, 3, 5, 7}，之后的位数只能是{1, 3, 7, 9}。这样最多只要8*（4^8）次判断素数。

三. 代码

AC代码，全部0s /* ID:maxkibb3 LANG:C++ PROG:sprime */ #include<cstdio> #include<cmath> using namespace std; int n, up = 1; int digit[2][10] = {{2, 3, 5, 7}, {1, 3, 7, 9}}; bool judge_prime(int num) { if(num == 1) return false; for(int i = 2; i <= sqrt(num); i++) { if(num % i == 0) return false; } return true; } void solve(int depth, int num) { if(depth == n) { printf("%d\n", num); return; } if(depth == 0) { for(int i = 0; i < 4; i++) { solve(depth + 1, digit[0][i]); } return; } for(int i = 0; i < 4; i++) { int _num = num * 10 + digit[1][i]; if(!judge_prime(_num)) continue; solve(depth + 1, _num); } } int main() { freopen("sprime.in", "r", stdin); freopen("sprime.out", "w", stdout); scanf("%d", &n); solve(0, 0); return 0; }