[搜索]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)

57

11
101
151131
181
353191
313
373
383

给定一个范围，输出在这个范围内（包含边界）的所有回文素数.
注意以后有回文数的情况，尽量先用DFS生成回文数表，然后在做处理，不要去枚举产生回文数.
这里打出回文数表以后先排序，然后二分查出上下界，把该范围中的所有素数输出即可。

代码：

/*
ID:yfr_1992
PROG:pprime
LANG:C++
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<cstdlib>
#include<algorithm>
#include<cmath>
using namespace std;
vector<int> plist;
int temp[10],sz;
void dfs(int cur,int limit){
if(cur>((limit-1)>>1)){
int sum = 0;
for(int i=0;i<limit;i++)sum = sum*10 + temp[i];
plist.push_back(sum);
return;
}
for(int i=0;i<10;i++){
temp[cur] = i,temp[limit-1-cur] = i;
dfs(cur+1,limit);
}
}
bool isPrime(int x){
for(int i=2;i<=(int)sqrt(x)+1;i++){
if((x%i)==0)return false;
}
return true;
}
int bs_lower(int x){
int l = 0, r = sz-1 , m ;
while(l<=r){
m = (l+r)>>1;
if(plist[m]>=x)r = m-1;
else l = m+1;
}
return l;
}
int bs_upper(int x){
int l = 0, r = sz-1 , m ;
while(l<=r){
m = (l+r)>>1;
if(plist[m]>x)r = m-1;
else l = m+1;
}
return r;
}
int main(){
freopen("pprime.in","r",stdin);
freopen("pprime.out","w",stdout);
for(int i=1;i<=8;i++)dfs(0,i);
sort(plist.begin(),plist.end());
sz = 1;
for(int i=1;i<(int)plist.size();i++){
if(plist[i]!=plist[i-1])plist[sz++] = plist[i];
}
int a,b;
scanf("%d%d",&a,&b);
int s = bs_lower(a), e = bs_upper(b);
for(int i=s;i<=e;i++){
if(isPrime(plist[i]))printf("%d\n",plist[i]);
}
return 0;
}