Rokua-Palindrome prime numbers-process functions and recursion

Title DescriptionDescription because 151 is both a prime and a palindrome number (from left to right and from right to left to see the same), so 151 is a palindrome prime.
Write a program to find all palindrome prime numbers between range [A, b] (5 <= a < <= 100,000,000) (100 million); input/output format input/output Input Format:
Line 1th: Two integers a and b.
output Format:
Output a list of palindrome prime numbers, one line at a. input and Output sample sample Input/output sample Test point # # Input Sample:5 500 Sample output:

5
7
11
101
131
151
181
191
313
353
373
383

DescriptionDescriptionhint 1:generate the palindromes and see if they is prime.
Tip 1: Find out all the palindrome numbers and then determine if they are prime numbers (primes).

Hint 2:generate palindromes by combining digits properly. You might need more than one of the loops like below.
Tip 2: To produce the correct palindrome number, you may need several loops like the one below.

The title translation comes from Nocow.
Usaco Training Section 1.5

Generate a palindrome number with a length of 5:

1  for(D1 =1; D1 <=9; d1+=2) {//only odd numbers are primes .2       for(D2 =0; D2 <=9; d2++) {3           for(D3 =0; D3 <=9; d3++) {4Palindrome =10000*d1 + +*D2 + -*d3 +Ten*d2 + D1;//(processing palindrome number ...)5          }6      }7}

Idea: This difficulty is a bit big, the scope unexpectedly has 100 million, too pit daddy! So I made a few simplifications to it:

① from the odd start, each time i=i+2;

② no even number of palindrome prime numbers, save a lot of time, it is really beneficial to mankind ah!

③ exclude multiples of 2, 5

Each function implementation process is detailed:

① Judging palindrome Number: Deposit An array to determine whether the previous sweep and backward sweep are consistent.

② determine the number of digits (whether the detection length is greater than 100 million): From the previous sweep, back to the number of cycles.

③ Judge Prime number (this is very important OH): I start from 3, each i=i+2, and then use that number to try to remove I, if just do, then is prime, return 1.

The code is as follows:

1#include <stdio.h>2#include <string.h>3 intHuiwen (intk);//Judging palindrome number4 intHwlength (intk);//calculate the length of a palindrome number5 intPrimeintk);//Judging Prime numbers6 intMain ()7 {8     inta,b,i,j;9scanf"%d%d",&a,&b);Ten      for(i=a;i<=b;i++) One     { A         if(i%2==0&&i!=2)//exclude multiples of 2 -             Continue; -         if(i%5==0&&i!=5)//exclude multiples of 5 the             Continue; -         if(Hwlength (i)%2==0&&i!= One)//the length of the palindrome number is even and not one -             Continue; -         if(Huiwen (i)! =1)//not a palindrome, just skip. +             Continue; -         if(Prime (i))//is prime +printf"%d\n", i); A     } at     return 0; - } - intHuiwen (intk) - { -     inta[Ten],i=0, J; -      while(k>0)//each bit is stored in an array and then judged in     { -a[i]=k%Ten; tok=k/Ten; +i++; -     } the      for(j=0; j<i;j++)//Double cycle judgment (front and rear sweep method) *         if(a[j]!=a[i-j-1])//No, return 0 $             return 0; Panax Notoginseng     return 1;//after the end, yes, return 1 - } the intHwlength (intk) + { A     inta[Ten],i=0; the      while(k>0) +     { -a[i]=k%Ten; $K/=Ten; $i++; -     } -     return(i);//I was cycled. the } - intPrimeintk)Wuyi { the     inti; -      for(i=3; i*i<=k;i=i+2)//This can reduce multiple cycles. Wu         if(k%i==0) -         { About             return 0; $         }            -     return 1; -}

Rokua-Palindrome prime numbers-process functions and recursion