Usaco--1.5prime palindromes

A property: The palindrome number of even digits is a multiple of 11 (except for 11 itself), so even if the bit is not considered, then we will construct the palindrome number of odd digits, and then judge whether this number is a prime line.

The code is as follows:

/*id:15674811lang:c++task:pprime*/#include <iostream>#include <cstdio>#include <cstring>#include <fstream>#include <cmath>using namespace STD;BOOLIs_prime (intN) {intk=sqrt(n) +0.5; for(intI=2; i<=k;i++)if(n%i==0)return 0;return 1;}/// structure to get the odd number of palindrome number within 10000000intGet (int*B) {intCnt=0;palindrome number of///1 bit        for(intI=1; i<=9; i++) b[cnt++]=i; b[cnt++]= One;/// even number of palindrome number is a multiple of 11, so do not consider        palindrome number of///3 bit         for(intI=1; i<=9; i++) for(intj=0; j<=9; j + +) {intd=i* -+j*Ten+i;            B[cnt++]=d; }palindrome number of///5 bit          for(intI=1; i<=9; i++) for(intj=0; j<=9; j + +) for(intk=0; k<=9; k++) {intD=10000*i+ +*j+ -*k+Ten*j+i;                 B[cnt++]=d; }palindrome number of///7 bit          for(intI=1; i<=9; i++) for(intj=0; j<=9; j + +) for(intk=0; k<=9; k++) for(intD=0;d <=9;d + +) {intd1=1000000*i+100000*j+10000*k+ +*d+ -*k+Ten*j+i;                  B[CNT++]=D1; }returnCNT;}intMain () {intb1[20000]; Ofstreamcout("Pprime.out"); IfstreamCin("Pprime.in");intCnt=get (B1);intb; while(Cin&GT;&GT;A&GT;&GT;B) { for(intI=0; i<cnt;i++)if(B1[i]>=a&&b1[i]<=b&&is_prime (B1[i]))cout<<b1[i]<<endl;Else if(B1[I]&GT;B) Break; }return 0;}

Usaco--1.5prime palindromes