ACM_Palindromic Prime Number Problem Solution Report: hdu 1431 Prime Palindromic Numbers

Palindromic prime

Time Limit: 2000/1000ms (Java/Others)

Problem Description:

Number 131 is a primary palindrome because it's a prime number and a palindrome (when read backwards, it's the same number). Write a program that counts the number 
of principal palindromes in the range of two provided numbers a and b (5 <= a <b <= 10^8) ; both a and b are considered to be in range. 
For example, the range 5-120 contains 3 principal palindromes (5, 7, 11, 101).

Input:

The input contains multiple cases. Each case contains two integers a and b.

Output:

For each case, output the answer to the question.

Sample Input:

5 11
5 120

Sample Output:

3
4 
Problem-solving ideas: This problem is the same as Hangdian hdu1431, link --> problem solution report: hdu 1431 prime number palindrome , but changed to the number of palindromic prime numbers in the output interval, also note that the input a may be is greater than b, so we need to judge.

 1 #include<bits/stdc++.h>
 2 #define maxn 9999999
 3 using namespace std;
 4 bool prime1[maxn];
 5 int cnt,a,b,num;
 6 bool prime2(int n)
 7 {
 8     int sum=0,tmp=n;
 9     while(tmp){
10         sum=sum*10+tmp%10;
11         tmp/=10;
12     }
13     if(sum==n)return true;
14     else return false;
15 }
16 int main()
17 {
18     memset(prime1,true,sizeof(prime1));
19     prime1[0]=prime1[1]=false;
20     for(int i=4;i<maxn;i+=2)prime1[i]=false;
21     for(int i=3;i<=3163;i+=2){
22         if(prime1[i]){
23             for(int j=i*i;j<maxn;j+=2*i)prime1[j]=false;
24         }
25     }
26     while(cin>>a>>b){
27         num=0;
28         for(int i=a;i<=b;i++){
29             if(i>=maxn)continue;
30             if(prime1[i]&&prime2(i))num++;
31         }
32         cout<<num<<endl;
33     }
34     return 0;
35 }