USACO Section 1.5 Prime Palindromes（回文数+素数）

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



题意：输入区间[a,b]求区间中的即是回文数又是素数的数。
分析：打出全部的回文数表，打出sqrt(b)中的素数表，然后再依次判断。
View Code 
  1 /*
  2 ID: dizzy_l1
  3 LANG: C++
  4 TASK: pprime
  5 */
  6 #include<iostream>
  7 #include<algorithm>
  8 #include<cstdio>
  9 #include<cmath>
 10 #include<stdlib.h>
 11 #include<cstring>
 12 
 13 using namespace std;
 14 
 15 int palind[20000],prime[10000],cnt1,cnt2;
 16 bool flag,isprime[10000];
 17 
 18 void dfs(int p,int n,char *s)
 19 {
 20     if(p>n) return ;
 21     if(p==n)
 22     {
 23         int q,i,k;
 24         char ts[10];
 25         for(i=0;i<10;i++)
 26             ts[i]=s[i];
 27         if(flag) q=p;
 28         else q=p-1;
 29         k=p+1;
 30         for(i=q;i>=1;i--)
 31             ts[k++]=s[i];
 32         palind[cnt1++]=atoi(ts);
 33         return ;
 34     }
 35     char c;
 36     for(c='0';c<='9';c++)
 37     {
 38         s[p+1]=c;
 39         dfs(p+1,n,s);
 40     }
 41 }
 42 
 43 void Init_palind(int n)
 44 {
 45     int i,j;
 46     char s[10],c;
 47     cnt1=0;
 48     if(n==9) n--;
 49     for(i=1;i<=n;i++)
 50     {
 51         memset(s,'\0',sizeof(s));s[0]='0';
 52         if(i&1) flag=false;
 53         else flag=true;
 54         j=(i+1)/2;
 55         for(c='1';c<='9';c++)
 56         {
 57             if(i==1)
 58             {
 59                 palind[cnt1++]=c-'0';
 60             }
 61             else
 62             {
 63                 s[1]=c;
 64                 dfs(1,j,s);
 65             }
 66         }
 67     }
 68    // for(i=0;i<cnt1;i++)
 69    //     printf("%d ",palind[i]);
 70 }
 71 
 72 void Init_prime(int n)
 73 {
 74     int i,j,m;
 75     cnt2=0;
 76     m=sqrt(n);
 77     memset(isprime,true,sizeof(isprime));
 78     isprime[1]=false;
 79     for(i=2;i<=m;i++)
 80     {
 81         if(isprime[i])
 82         {
 83             prime[cnt2++]=i;
 84             for(j=i+i;j<=m;j=j+i)
 85             {
 86                 isprime[j]=false;
 87             }
 88         }
 89     }
 90   //  for(i=0;i<cnt2;i++)
 91   //      printf("%d ",prime[i]);
 92 }
 93 
 94 bool judge(int n)
 95 {
 96     int i;
 97     for(i=0;i<cnt2&&prime[i]*prime[i]<=n;i++)
 98     {
 99         if(n%prime[i]==0)
100             return false;
101     }
102     return true;
103 }
104 
105 void work(int a,int b)
106 {
107     int i,j,k;
108     i=0;j=cnt1-1;
109     while(palind[i]<a) i++;
110     while(palind[j]>b) j--;
111     for(k=i;k<=j;k++)
112     {
113         if(judge(palind[k]))
114         {
115             printf("%d\n",palind[k]);
116         }
117     }
118 }
119 
120 int main()
121 {
122     freopen("pprime.in","r",stdin);
123     freopen("pprime.out","w",stdout);
124     int a,b;
125     char ca[15],cb[15];
126     while(scanf("%s %s",ca,cb)==2)
127     {
128         a=atoi(ca);
129         b=atoi(cb);
130         Init_palind(strlen(cb));
131         Init_prime(b);
132         work(a,b);
133     }
134     return 0;
135 }

 



posted @ 2012-08-29 21:20  mtry  阅读(1028)  评论(0)  编辑  收藏  举报