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
1 /* 2 ID:spcjv51 3 PROG:pprime 4 LANG:C 5 */ 6 #include<stdio.h> 7 #include<math.h> 8 long f[15]; 9 long a,b; 10 int is_prime(long n) 11 { 12 long i; 13 if(n<2) return 0; 14 for(i=2; i<=sqrt(n); i++) 15 if(n%i==0) return 0; 16 return 1; 17 } 18 int main(void) 19 { 20 freopen("pprime.in","r",stdin); 21 freopen("pprime.out","w",stdout); 22 long i,d1,d2,d3,d4,palindrome; 23 scanf("%ld%ld",&a,&b); 24 for(i=a; i<10; i++) 25 if(is_prime(i)) printf("%ld\n",i); 26 if(a<11&&b>11) printf("%d\n",11); 27 for (d1 = 1; d1 <= 9; d1+=2) 28 for (d2 = 0; d2 <= 9; d2++) 29 { 30 palindrome = 100*d1 + 10*d2 + d1; 31 if(palindrome>=a&&palindrome<=b&&is_prime(palindrome)) 32 printf("%ld\n",palindrome); 33 34 35 } 36 for (d1 = 1; d1 <= 9; d1+=2) 37 for (d2 = 0; d2 <= 9; d2++) 38 for(d3=0; d3<=9; d3++) 39 { 40 palindrome =10000*d1+1000*d2+100*d3 + 10*d2 + d1; 41 if(palindrome>=a&&palindrome<=b&&is_prime(palindrome)) 42 printf("%ld\n",palindrome); 43 44 } 45 for (d1 = 1; d1 <= 9; d1+=2) 46 for (d2 = 0; d2 <= 9; d2++) 47 for(d3=0; d3<=9; d3++) 48 for(d4=0; d4<=9; d4++) 49 { 50 palindrome =1000000*d1+100000*d2+10000*d3+1000*d4+100*d3 + 10*d2 + d1; 51 if(palindrome>=a&&palindrome<=b&&is_prime(palindrome)) 52 printf("%ld\n",palindrome); 53 54 } 55 return 0; 56 }