USACO1.5.2--Prime Palindromes

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!)

Prime Palindromes: Hint 1

Generate the palindromes and see if they are prime.

Prime Palindromes: 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 ...
}
}
}
题解：枚举。可以先生成回文数然后再判断是否是质数。我第一次是用最朴素的方法，一亿果断超时了。生成回文数的时候有个小技巧，也就是Hint 2提到的，因为回文数是对称的，所以我么只要枚举回文数的一半就行，然后再倒转一下就OK了。并且偶数长度的回文数全部能整除11，所以我么只需枚举奇数长度的回文数就可以了，唯一特殊的偶数长度的回文数就是11。这样的话最大的数据枚举量也就是5*9*9*9=3645。比直接枚举一亿，少了好多！！！

View Code

/*
ID:spcjv51
PROG:pprime
LANG:C
*/
#include<stdio.h>
#include<math.h>
long f[15];
long a,b;
int is_prime(long n)
{
long i;
if(n<2) return 0;
for(i=2; i<=sqrt(n); i++)
if(n%i==0) return 0;
return 1;
}
int main(void)
{
freopen("pprime.in","r",stdin);
freopen("pprime.out","w",stdout);
long i,d1,d2,d3,d4,palindrome;
scanf("%ld%ld",&a,&b);
for(i=a; i<10; i++)
if(is_prime(i)) printf("%ld\n",i);
if(a<11&&b>11) printf("%d\n",11);
for (d1 = 1; d1 <= 9; d1+=2)
for (d2 = 0; d2 <= 9; d2++)
{
palindrome = 100*d1 + 10*d2 + d1;
if(palindrome>=a&&palindrome<=b&&is_prime(palindrome))
printf("%ld\n",palindrome);

}
for (d1 = 1; d1 <= 9; d1+=2)
for (d2 = 0; d2 <= 9; d2++)
for(d3=0; d3<=9; d3++)
{
palindrome =10000*d1+1000*d2+100*d3 + 10*d2 + d1;
if(palindrome>=a&&palindrome<=b&&is_prime(palindrome))
printf("%ld\n",palindrome);

}
for (d1 = 1; d1 <= 9; d1+=2)
for (d2 = 0; d2 <= 9; d2++)
for(d3=0; d3<=9; d3++)
for(d4=0; d4<=9; d4++)
{
palindrome =1000000*d1+100000*d2+10000*d3+1000*d4+100*d3 + 10*d2 + d1;
if(palindrome>=a&&palindrome<=b&&is_prime(palindrome))
printf("%ld\n",palindrome);

}
return 0;
}