Hoj 1004 Prime Palindromes
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本题练习穷举。
题目链接:http://acm.hit.edu.cn/hoj/problem/view?id=1004
本题以前做过,之所以记录本题,是因为看到一种生成回文数的有效方法,套用此模板可以快速生成1 - 10^9范围内的回文数表。虽然看似繁杂,但是却非常有效。
本题是求一定范围内的所有回文素数。
#include <iostream>#include <stdio.h>#include <string.h>#include <math.h>#include <algorithm>using namespace std;int main(){#ifndef ONLINE_JUDGE freopen("in.txt","r",stdin);#endif int a,b; scanf("%d %d",&a,&b); int Palind[110000],pnum=0; int n,m,k,p,q; //求所有的回文数 for(n=1; n<=9; n++) Palind[pnum++]=n; for(n=1; n<=9; n++) Palind[pnum++]=11*n; for(n=1; n<=9; n++) for(m=0; m<=9; m++) Palind[pnum++]=101*n+10*m; for(n=1; n<=9; n++) for(m=0; m<=9; m++) Palind[pnum++]=1001*n+110*m; for(n=1; n<=9; n++) for(m=0; m<=9; m++) for(k=0; k<=9; k++) Palind[pnum++]=10001*n+1010*m+100*k; for(n=1; n<=9; n++) for(m=0; m<=9; m++) for(k=0; k<=9; k++) Palind[pnum++]=100001*n+10010*m+1100*k; for(n=1; n<=9; n++) for(m=0; m<=9; m++) for(k=0; k<=9; k++) for(p=0; p<=9; p++) Palind[pnum++]=1000001*n+100010*m+10100*k+1000*p; for(n=1; n<=9; n++) for(m=0; m<=9; m++) for(k=0; k<=9; k++) for(p=0; p<=9; p++) Palind[pnum++]=10000001*n+1000010*m+100100*k+11000*p; for(n=1; n<=9; n++) for(m=0; m<=9; m++) for(k=0; k<=9; k++) for(p=0; p<=9; p++) for(q=0; q<=9; q++) Palind[pnum++]=100000001*n+10000010*m+1000100*k+101000*p+10000*q; //判断素数 for(int i=0;i<pnum;i++) { if(Palind[i]>b) { break; } int q = sqrt(Palind[i]) + 1; int flag = 0; for(int j = 2;j<q;j++) { if(Palind[i]%j == 0) { flag = 1; break; } } if(flag == 0 && Palind[i]>=a) { printf("%d\n",Palind[i]); } } return 0;}
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