Emirp
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EmirpTime Limit: 5000 MSMemory Limit: 100000 KTotal Submit: 143(59 users)Total Accepted: 61(55 users)Rating: Special Judge: NoDescription
An emirp (prime spelled backwards) is a prime number that results in a different prime when its decimal digits are reversed.
The first five emirps are 13, 17, 31, 37, 71
Now Kim want to know the kth emirp. Help him.
InputThe first line is an integer T, describes the number of tests. Then T tests.
In each test, one line an integer k.
OutputFor each test, output the kth emirp.
Sample Input3123Sample Output
131731Hint
T<=10000
k<=1000
思路:就是找大于十的数使其本身和其反置都为素数,并且本身和其反置数不能相同。
#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>#include<cmath>using namespace std;int a[1000000];int sushu(int n){ int i,j; int k; k=sqrt(n); for(i=2;i<=k;i++) { if(n%i==0) break; } if(i>k) return 1; else return 0;}int main(){ int i,j,k,k1,b,c,d,t,sum; for(i=13,j=1; i<=70529; i++) { k=0; t=i; while(t>0) { k=k*10+t%10; t=t/10; } k1=i; if(sushu(k)==1&&sushu(k1)==1&&k1!=k) { a[j]=k1; j++; } } int n,m; scanf("%d",&n); while(n--) { scanf("%d",&m); printf("%d\n",a[m]); }}
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