Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9266 Accepted Submission(s): 2167
所以;除了11是回文素数以外,其他偶数位回文数均可被11整除并且是合数,
然后就枚举1位3位5位7位9位数的数,然后判断是否为素数即可!!!
本人不才只能想到这菜鸟的思路了,不过浅显易懂啊!o(∩_∩)o 哈哈!
ps:http://acm.hdu.edu.cn/showproblem.php?pid=1431
#include<stdio.h> __int64 a,b; bool judge(__int64 x)//判断素数 { for(int i=2;i*i<=x;i++) if(x%i==0) return false; return true; } void solve() { __int64 ans,i,j,k,m,r; if(a<=11) //1位素数和11 { for(i=a;i<=11;i++) { if(i>b) return ; if(judge(i)) printf("%I64d\n",i); } } if(b>100) //3位数时 { for(i=1;i<=9;i+=2) for(j=0;j<=9;j++) { ans = i*101+j*10; if(ans>=a&&ans<=b&&judge(ans)) printf("%I64d\n",ans); if(ans>=b) return; } } if(b>10000) //5位数时 { for(i=1;i<=9;i+=2) for(j=0;j<=9;j++) for(k=0;k<=9;k++) { ans = i*10001+j*1010+k*100; if(ans>=a&&ans<=b&&judge(ans)) printf("%I64d\n",ans); if(ans>=b) return; } } if(b>1000000) //7位数时 { for(i=1;i<=9;i+=2) for(j=0;j<=9;j++) for(k=0;k<=9;k++) for(m=0;m<=9;m++) { ans = i*1000001+j*100010+k*10100+m*1000; if(ans>=a&&ans<=b&&judge(ans)) printf("%I64d\n",ans); if(ans>=b) return; } } if(b>100000000) //9位数时 { for(i=1;i<=9;i+=2) for(j=0;j<=9;j++) for(k=0;k<=9;k++) for(m=0;m<=9;m++) for(r=0;r<=9;r++) { ans = i*100000001+j*10000010+k*1000100+m*101000+r*10000; if(ans>=a&&ans<=b&&judge(ans)) printf("%I64d\n",ans); if(ans>=b) return; } } } int main() { while(scanf("%I64d%I64d",&a,&b)!=EOF) { if(a>b) { a=a^b; b=a^b; a=a^b; } solve(); printf("\n"); } return 0; }
吐槽:寒假要在家编程真的是。。。。。无力啊!早上又起不来,,,然后关键是心不静啊!(┬_┬)
原文:http://www.cnblogs.com/yuyixingkong/p/3529690.html