USACO Prime Palindromes(dfs)
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题目请点我
题解:
这道题看起来是一道很简单的题目,但是数据量很大。如果暴力判断会超时,预处理筛法标记素数则会超内存。最后看了网上的解法才知道要逆推,生成回文数,这还是要第一次遇到。并且很重要的一点是偶数位的回文数都能被11整除,所以只需要生成奇数位的,偶数位回文素数符合情况的值可能有11一个。挺考验思维的,重要的是do mathmatics first.
代码实现:
/*ID: eashionLANG: C++TASK: pprime*/#include <iostream>#include <cstdio>#include <cmath>#include <cstdlib>#include <cstring>#include <algorithm>#define MAX 10010using namespace std;int a,b;int up;int pos;int num[12];int save[MAX];void dfs(int len);bool if_prime(int x);int get_Palindrome(int len);int main(){ freopen("pprime.in","r",stdin); freopen("pprime.out","w",stdout); while( scanf("%d%d",&a,&b) != EOF ){ pos = 0; memset(save,0,sizeof(save)); memset(num,0,sizeof(num)); dfs(0); if( 11 >= a && 11 <= b ){ save[pos++] = 11; } sort(save,save+pos); for( int i = 0; i < pos; i++ ){ printf("%d\n",save[i]); } } return 0;}////void pre(){// for( int i = 0; i < MAX+1; i++ ){// is_prime[i] = true;// }// is_prime[0] = is_prime[1] = false;// for( int i = 2; i < MAX+1; i++ ){// if( is_prime[i] ){// for( int j = 2*i; j <= MAX; j+=i ){// is_prime[j] = false;// }// }// }//}void dfs(int len){ if( len > 5 ){ return ; } int tmp = get_Palindrome(len); if( tmp > b ){ return; } if( tmp >= a && tmp <= b && if_prime(tmp) ){ save[pos++] = tmp; } for( int i = 0; i < 10; i++ ){ if( len == 0 && i == 0 ){ continue; } num[len] = i; dfs(len+1); } return ;}int get_Palindrome(int len){ int tmp[12]; int up_B = 2*len-1; for( int i = 0; i < up_B; i++ ){ if( i < len ){ tmp[i] = num[i]; } else{ tmp[i] = num[up_B-1-i]; } } int Pnum = 0; for( int i = up_B-1; i >= 0; i-- ){ Pnum *= 10; Pnum += tmp[i]; } return Pnum;}bool if_prime(int x){ int up_B = sqrt(x); for( int i = 2; i <= up_B; i++ ){ if( x%i == 0 ){ return false; } } return true;}
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