Codeforces Round #549 (Div. 2) B. Nirvana

[題目描述]

B. Nirvana

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

Kurt reaches nirvana when he finds the product of all the digits of some positive integer. Greater value of the product makes the nirvana deeper.

Help Kurt find the maximum possible product of digits among all integers from 1 to n.

Input

The only input line contains the integer n (1≤n≤2⋅109).

Output

Print the maximum product of digits among all integers from 1 to n.

Examples

input

390

output

216

input

7

output

7

input

1000000000

output

387420489

Note

In the first example the maximum product is achieved for 389389 (the product of digits is 3⋅8⋅9=2163⋅8⋅9=216).

In the second example the maximum product is achieved for 77 (the product of digits is 77).

In the third example the maximum product is achieved for 999999999999999999 (the product of digits is 99=38742048999=387420489).

[解法]

盡量將每一位變為9，或者向前退位來變成9，並且前一位減1.

[代碼（AC）]

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <string>
 4 #include <iostream>
 5 #include <iostream>
 6 #include <iostream>
 7 #include <iostream>
 8 #include <iostream>
 9 using namespace std;
10 int ans(int n){
11     if(n==0)return 1;
12     if(n<10)return n;
13     return max(ans(n/10)*(n%10),ans(n/10-1)*9);
14 }
15 int main() 
16 {
17     int n;
18     scanf("%d", &n);
19     printf("%d", ans(n));
20 } 

2019-04-04 18:05:50

[題解] Codeforces Round #549 (Div. 2) B. Nirvana