codeforces 569C C. Primes or Palindromes?(素数筛+dp)

题目链接：C. Primes or Palindromes?time limit per test
3 secondsmemory limit per test
256 megabytesinput
standard inputoutput
standard outputRikhail Mubinchik believes that the current definition of prime numbers is obsolete as they are too complex and unpredictable. A palindromic number is another matter. It is aesthetically pleasing, and it has a number of remarkable properties. Help Rikhail to convince the scientific community in this!Let us remind you that a number is called prime if it is integer larger than one, and is not divisible by any positive integer other than itself and one.Rikhail calls a number a palindromic if it is integer, positive, and its decimal representation without leading zeros is a palindrome, i.e. reads the same from left to right and right to left.One problem with prime numbers is that there are too many of them. Let's introduce the following notation: π(n) — the number of primes no larger than n, rub(n) — the number of palindromic numbers no larger than n. Rikhail wants to prove that there are a lot more primes than palindromic ones.He asked you to solve the following problem: for a given value of the coefficient A find the maximum n, such that π(n) ≤ A·rub(n).Input
The input consists of two positive integers p, q, the numerator and denominator of the fraction that is the value of A (

,

).Output
If such maximum number exists, then print it. Otherwise, print "Palindromic tree is better than splay tree" (without the quotes).Examplesinput

1 1

output

40

input

1 42

output

1

input

6 4

output

172

题意：

问满足pi
/rub
<=p/q的最大的n是多少;

思路：

pi[i]和rub[i]都随着i的增大而增大,且pi[i]/rub[i]的值也随着增大,(小于10的数特殊);p/q给有范围,可以算一下大约1200000时pi[i]/rub[i]已经大约42了;所以暴力找到那个最大的n;

AC代码：

/*2014300227    569C - 28    GNU C++11    Accepted    61 ms    14092 KB*/
#include <bits/stdc++.h>
using namespace std;
const int N=12e5+4;
typedef long long ll;
const double PI=acos(-1.0);
int p,q,pi
,vis
,rub
;
void get_pi()//素数筛+dp得到pi[i]
{
memset(pi,0,sizeof(pi));
pi[1]=0;
for(int i=2;i<N;i++)
{
if(!pi[i])
{
for(int j=2;j*i<N;j++)
{
pi[i*j]=1;
}
pi[i]=pi[i-1]+1;
}
else pi[i]=pi[i-1];
}
}
int is_pal(int x)//判断一个数是不是回文数;
{
int s=0,y=x;
while(y)
{
s*=10;
s+=y%10;
y/=10;
}
if(s==x)return 1;
return 0;
}
void get_rub()
{
rub[0]=0;
for(int i=1;i<N;i++)
{
if(is_pal(i))rub[i]=rub[i-1]+1;
else rub[i]=rub[i-1];
}
}
int check(int x)
{
if(pi[x]*q<=p*rub[x])return 1;
return 0;

}
int get_ans()
{
int ans=0;
for(int i=1;i<N;i++)
{
if(check(i))ans=i;
}
if(ans==0)printf("Palindromic tree is better than splay tree\n");
else printf("%d\n",ans);
}
int main()
{
get_pi();
get_rub();
//cout<<pi[1200000]*1.0/(rub[1200000]*1.0);
scanf("%d%d",&p,&q);
get_ans();

return 0;
}