Prime Palindromes

Problem：

Prime Palindromes

The number 151 is a prime palindrome because it is both a prime number and a palindrome (it is the same number when read forward as backward). Write a program that finds all prime palindromes in the range of two supplied numbers a and b (5 <= a < b <= 100,000,000); both a and b are considered to be within the range .

PROGRAM NAME: pprime

INPUT FORMAT

Line 1:

Two integers, a and b

SAMPLE INPUT (file pprime.in)

5 500

OUTPUT FORMAT

The list of palindromic primes in numerical order, one per line.

SAMPLE OUTPUT (file pprime.out)

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

HINTS (use them carefully!)
 
Hint 1
Generate the palindromes and see if they are prime. 
 
Hint 2
Generate palindromes by combining digits properly. You might need more than one of the loops like below. 
/* generate five digit palindrome: */
for (d1 = 1; d1 <= 9; d1+=2) {	/* only odd; evens aren't so prime */
    for (d2 = 0; d2 <= 9; d2++) {
        for (d3 = 0; d3 <= 9; d3++) {
	    palindrome = 10000*d1 + 1000*d2 +100*d3 + 10*d2 + d1;
	    ... deal with palindrome ...
	}
    }
}

My Answer：

 

/* ID:iby071 LANG:C++ TASK:pprime */ #include<iostream> #include<fstream> #include<cmath> using namespace std; ifstream fin("pprime.in"); ofstream fout("pprime.out"); bool isPrime(long int n) //判断是否为质数 { double k=sqrt(static_cast<double>(n)); for(int i=2;i<=k;++i) if(n%i==0) return false; return true; } void pal1(long int a,long int b) { for(int d1=1;d1<=9;d1+=2) if(d1>=a && d1<=b && isPrime(d1)) fout<<d1<<endl; } void pal2(long int a,long int b) { if(11>=a && 11<=b) fout<<"11"<<endl; } void pal3(long int a,long int b) { for(int d1=1;d1<=9;d1+=2) for(int d2=0;d2<=9;++d2) { int panlindrome=d1*100+d2*10+d1; if(panlindrome>=a && panlindrome<=b && isPrime(panlindrome)) fout<<panlindrome<<endl; } } void pal5(long int a,long int b) { for(int d1=1;d1<=9;d1+=2) for(int d2=0;d2<=9;++d2) for(int d3=0;d3<=9;++d3) { int panlindrome=d1*10000+d2*1000+d3*100+d2*10+d1; if(panlindrome>=a && panlindrome<=b && isPrime(panlindrome)) fout<<panlindrome<<endl; } } void pal7(long int a,long int b) { for(int d1=1;d1<=9;d1+=2) for(int d2=0;d2<=9;++d2) for(int d3=0;d3<=9;++d3) for(int d4=0;d4<=9;++d4) { int panlindrome=d1*1000000+d2*100000+d3*10000+d4*1000+d3*100+d2*10+d1; if(panlindrome>=a && panlindrome<=b && isPrime(panlindrome)) fout<<panlindrome<<endl; } } void pal9(long int a,long int b) { for(int d1=1;d1<=9;d1+=2) for(int d2=0;d2<=9;++d2) for(int d3=0;d3<=9;++d3) for(int d4=0;d4<=9;++d4) for(int d5=0;d5<=9;++d5) { int panlindrome=d1*100000000+d2*10000000+d3*1000000+d4*100000+d5*10000+d4*1000+d3*100+d2*10+d1; if(panlindrome>=a && panlindrome<=b && isPrime(panlindrome)) fout<<panlindrome<<endl; } } void defaultFunc(long int a,long int b){} //为了把所有生成不同位数palindrome的函数做成一个数组而定义的空函数 int main() { long int a,b,tmp; int digita=1,digitb=1; void (*palFunc[10])(long int,long int)={defaultFunc,pal1,pal2,pal3,defaultFunc,pal5,defaultFunc,pal7,defaultFunc,pal9}; //编入数组，便于调用 fin>>a>>b; tmp=a/10; while(tmp!=0) { ++digita; tmp/=10; } tmp=b/10; while(tmp!=0) { ++digitb; tmp/=10; } for(int i=digita;i<=digitb;++i) palFunc[i](a,b); return 0; }

 

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My Gain:

一个重要的定理：

质数prime：位数为偶数位panlindrome的质数有且只有11。

由此可以少做很多次不必要的isPrime检查。