1.5 超级肋骨质数

Superprime Rib

Butchering Farmer John's cows always yields the best prime rib. Youcan tell prime ribs by looking at the digits lovingly stamped acrossthem, one by one, by FJ and the USDA. Farmer John ensures that apurchaser of his prime ribs gets really prime ribs because when slicedfrom 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 areprime; the two ribs 73 are prime, and, of course, the last rib, 7, isprime. The number 7331 is called a superprime of length 4.

Write a program that accepts a number N 1 <=N<=8 of ribs andprints 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 perline.

SAMPLE OUTPUT (file sprime.out)

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

/*ID: xdzhbin1LANG: C++TASK: sprime*///超级肋骨质数，如7331，7是质数，73是质数，733也是质数，7331还是质数//找出n位的所有超级肋骨质数，这样的数是由一个质数乘以10加上一个个位数产生的#include <vector>#include <fstream>#include <iostream>using namespace std;//10000以下质数列表void putprime(vector<long>& v){v.push_back(2);int i,j;bool isprime;for(i=3; i<10000; i++){isprime = true;for (j=0; j<v.size(); j++){if (i%v[j]==0){isprime = false;break;}}if (isprime){v.push_back(i);}}}//判断num是否为质数bool isprime(long num, vector<long>& v){int i=0;while(v[i]*v[i]<=num){if(num % v[i] == 0)return false;i++;}return true;}int main(){ifstream in("sprime.in",ios::in);ofstream out("sprime.out",ios::out);int n;in>>n;//读入位数vector<long> vprime; putprime(vprime); //产生质数列表typedef vector<long> array;vector<array> v;v.resize(8);v[0].push_back(2);//放入一位的质数v[0].push_back(3);v[0].push_back(5);v[0].push_back(7);int i,j,k;for(i=1; i<8; i++){for(j=0; j<v[i-1].size(); j++){for(k=0; k<10; k++){int temp=v[i-1][j]*10 + k;if (isprime(temp,vprime)){v[i].push_back(temp);}}}}for (i=0; i<v[n-1].size(); i++){out<<v[n-1][i]<<endl;}return 0;}