Codeforces Round #315 (Div. 2) C. Primes or Palindromes? 素数表 回文数

C. Primes or Palindromes?

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Rikhail Mubinchik believes that the current definition of prime numbers is obsolete as they are too complex and unpredictable.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).

Sample test(s)

input

1 1

output

40

input

1 42

output

1

input

6 4

output

172

由于，p/q很小，直接打素数表，统计一下，就可以了。求回文串，直接用暴力求即可。

#define N 1200050
#define M 100005
#define maxn 205
#define MOD 1000000000000000007
int n,np[N],hp[N],num[30],nn;
bool pri[N];
int isHp(int x){
    int xx = x;
    nn = 0;
    while(x){
        num[nn++] = x % 10;
        x/=10;
    }
    for(int i = 0;i<nn/2;i++){
        if(num[i] != num[nn - 1 - i]) return 0;
    }
    //printf("%d ",xx);
    return 1;
}
void init(){
    fill(pri,true);
    for(int i =2;i<N;i++){
        if(pri[i]){
            for(int j = i + i;j<N;j+=i){
                pri[j] = false;
            }
        }
    }
    np[0] = np[1] = 0;
    for(int i = 2;i<N;i++){
        np[i] = np[i-1];
        if(pri[i]) np[i]++;
    }
    hp[0] = 0;
    for(int i = 1;i<N;i++){
        hp[i] = hp[i-1] + isHp(i);
    }
    //for(int i =0;i<100;i++)
    //int i  = N - 1;
    //{
    //    printf("%d %d %f\n",np[i],hp[i],(double)np[i]/(double)hp[i]);
    //}
}
int main()
{
    init();
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int p,q;
     //printf("%d %d \n",np[0],hp[0]);
     while(S2(p,q)!=EOF)
    {

        bool flag = true;
        for(int i = N - 1;i >= 0 && flag;i--){
            if(q * np[i] <= p * hp[i]){
                flag = false;
                printf("%d\n",i);
            }
        }
        if(flag){
            printf("Palindromic tree is better than splay tree\n");
        }
    }
    //fclose(stdin);
    //fclose(stdout);
    return 0;
}

大规模