About the most number icodelab

description

For a positive integer, if all the number is less than a divisor of the number are less than the number itself about a number, I think this number is exactly what we want.

Entry

Enter a positive integer X.

Export

X a is not greater than the output of satisfying the above requirements and the maximum number a.

Sample input 1

1000

Sample Output 1

840

prompt

For 10% of the data, 1 <= n <= 1,000. For 40% of the data, 1 <= n <= 1,000,000. To 100% of the data, 1 <= n <= 2,000,000,000.

Code (Why did not write it borrowed ideas teacher code, detailed notes?):

#include <cstdio> 
#include <the iostream> 
the using namespace STD; 
typedef Long Long LL; 
int Prime [12 is] = {0, 2,. 3,. 5,. 7,. 11, 13 is,. 17,. 19, 23 is, 29}; 
/ / 2 * 3 * 5 * 7 * 11 * 13 * 17 * 19 * 23 * 29> n, thus taking into account only 29 to 
LL n-, BestSum, BestNum; 
// current number of the num go, followed by the first k prime numbers, the number of divisor num SUM, 
// the upper limit of the number k primes to limit 
void Solve (num LL, LL SUM, limit LL, LL k) { 
    IF (SUM> BestSum) {// found more divisor of the number, the more stored divisor of the number of natural 
        BestSum = sum; // save the number 
        BestNum = num; // save the number itself 
    } else if (sum == BestSum && num <BestNum) {/ when the same / divisor number, taking the fractional 
        BestNum NUM =; 
    } 
    for (int i =. 1; i <= limit; i ++) {// i th prime number k takes 
        num * = prime [k]; 
        if (num> n) // num more than n , apparently you do not need to go to find
            return; 
        // second parameter sum * (1 + i) represents the number of the current number of about num, reference is prime decomposition theorem 
        // third parameter is i, the latter represents a number of power can not exceed the number in front of power 
        Solve (num, SUM * (. 1 + I), I, K +. 1); 
    } 
} 
int main () {     
    CIN n->>; 
    // Solve (num, SUM, limit, K) 
    // num current went this number is the number of divisor num sum, followed by the first k prime numbers, the upper limit of the number of k primes limit     
    Solve (. 1,. 1, 30,. 1); 
    COUT << BestNum; 
    return 0; 
}