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  • USACO 1.5 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

    ——————————————————————题解
    就是枚举回文数然后筛,写了一个logn的筛法(Miller Rabin),结果乘法没开longlong导致wa了一次
    枚举素数我手敲了八个循环……后来ac看analysis……只要10-9999然后翻转就行orz我感觉自己真是个辣鸡
    而且这位小哥以神一般的数学直觉便判断出任何二倍长的回文数一定是11的倍数,那么我们就不用考虑偶数个的了。小学数学老师你还要我吗要我吗……
    这轻描淡写的一句话顶上了我30++行代码orz
    那个版权不敢盗,所以不粘别人代码,贴上我易读的代码……以及写的很丑的Miller Rabin(check函数)
      1 /*
      2 PROB: pprime
      3 LANG: C++
      4 ID: jiaqi si
      5 */
      6 #include <iostream>
      7 #include <string.h>
      8 #include <cstdlib>
      9 #include <cstdio>
     10 #include <algorithm>
     11 #include <cstring>
     12 #include <vector>
     13 #include <ctime>
     14 #define ivory
     15 #define mo  1000000007 
     16 #define siji(i,x,y) for(int i=(x);i<=(y);i++)
     17 #define gongzi(j,x,y) for(int j=(x);j>=(y);j--)
     18 #define xiaosiji(i,x,y) for(int i=(x);i<(y);i++)
     19 #define sigongzi(j,x,y) for(int j=(x);j>(y);j--)
     20 #define pii pair<int,int>
     21 #define fi first
     22 #define se second
     23 #define mo 1000000007
     24 using namespace std;
     25 int ans[500005],cnt;
     26 int a,b;
     27 int mpow(int c,int d,int k) {
     28     if(d==1) {
     29         return c%k;
     30     }
     31     else {
     32         int tmp=mpow(c,d/2,k)%k;
     33         if(d&1) {
     34             return 1LL*tmp*tmp%k*c%k;
     35         }
     36         else {
     37             return 1LL*tmp*tmp%k;
     38         }
     39     }
     40 }
     41 bool _check(int v,int s,int k) {
     42     if(v==1) return true;
     43     siji(i,1,s) {
     44         if(1LL*v*v%k==1) {
     45             if(v%k==k-1){return true;}
     46             else return false;
     47         }
     48         v=1LL*v*v%k;
     49     }
     50     return false;
     51 }
     52 bool check(int k) {
     53     int tmp=k-1;
     54     int s=1;
     55     while(tmp%(1<<s)==0) ++s;
     56     --s;
     57     int d=tmp/(1<<s);
     58     int tmp1=mpow(2,d,k);
     59     int tmp2=mpow(7,d,k);
     60     int tmp3=mpow(61,d,k);
     61     bool f=1;
     62     if(!_check(tmp1,s,k)) {f=0;}
     63     else if(k>7 && (!_check(tmp2,s,k))) {f=0;}
     64     else if(k>61 && (!_check(tmp3,s,k))) {f=0;}
     65     return f;
     66 }
     67 int binary(int k){
     68     int l=1,r=cnt;
     69     while(l<r) {
     70         int mid=(l+r+1)>>1;
     71         if(ans[mid]>=k) r=mid-1;
     72         else l=mid;
     73     }
     74     return l;
     75 }
     76 int main() {
     77 #ifdef ivory
     78     freopen("pprime.in","r",stdin);
     79     freopen("pprime.out","w",stdout);
     80 #else 
     81     //freopen("f1.in","r",stdin);
     82 #endif
     83     ans[++cnt]=2;
     84     siji(i,3,9) {
     85         if(i%2!=0 && check(i)) ans[++cnt]=i;
     86     }
     87     siji(i,1,9) {
     88         if(i%2==0)continue;
     89         int tmp=i*10+i;
     90         if(check(tmp)) ans[++cnt]=tmp;
     91     }
     92     siji(i,1,9) {
     93         if(i%2==0) continue;
     94         siji(j,0,9) {
     95             int tmp=i*100+j*10+i;
     96             if(check(tmp)) ans[++cnt]=tmp;
     97         }
     98     }
     99     siji(i,1,9) {
    100         if(i%2==0) continue;
    101         siji(j,0,9) {
    102             int tmp=i*1000+j*100+j*10+i;
    103             if(check(tmp)) ans[++cnt]=tmp;
    104         }
    105     }
    106     siji(i,1,9) {
    107         if(i%2==0) continue;
    108         siji(j,0,9) {
    109             siji(k,0,9) {
    110                 int tmp=i*10000+j*1000+k*100+j*10+i;
    111                 if(check(tmp)) ans[++cnt]=tmp;
    112             }
    113         }
    114     }
    115     siji(i,1,9) {
    116         if(i%2==0) continue;
    117         siji(j,0,9) {
    118             siji(k,0,9) {
    119                 int tmp=i*100000+j*10000+k*1000+k*100+j*10+i;
    120                 if(check(tmp)) ans[++cnt]=tmp;
    121             }
    122         }
    123     }
    124     siji(i,1,9) {
    125         if(i%2==0) continue;
    126         siji(j,0,9) {
    127             siji(k,0,9) {
    128                 siji(h,0,9) {
    129                     int tmp=i*1000000+j*100000+k*10000+h*1000+k*100+j*10+i;
    130                     if(check(tmp)) ans[++cnt]=tmp;
    131                 }
    132                 
    133             }
    134         }
    135     }
    136     siji(i,1,9) {
    137         if(i%2==0) continue;
    138         siji(j,0,9) {
    139             siji(k,0,9) {
    140                 siji(h,0,9) {
    141                     int tmp=i*10000000+j*1000000+k*100000+h*10000+h*1000+k*100+j*10+i;
    142                     if(check(tmp)) ans[++cnt]=tmp;
    143                 }
    144             }
    145         }
    146     }
    147     scanf("%d%d",&a,&b);
    148     int il=binary(a);
    149     int ir=binary(b);
    150     if(ans[ir+1]==b) ir++;
    151     siji(i,il+1,ir) {
    152         printf("%d
    ",ans[i]);
    153     }
    154 }
    
    
    
     
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  • 原文地址:https://www.cnblogs.com/ivorysi/p/5818707.html
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