1143 B. Nirvana

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<string>
#include<cmath>
#include<cstring>
#include<cstdlib>
using namespace std;
typedef long long ll;

ll getvalue(string s) {
    ll sum = 1;
    for (int i = 0; i < s.size(); i++) {
        sum *= s[i] - '0';
    }
    return sum;
}
ll calc(string s) {
    string q1 = s;
    q1[0]--;
    for (int i = 1; i < q1.size(); i++)q1[i] = '9';
    return max({
        getvalue(s),
        getvalue(q1),
        getvalue(q1.substr(1)),
        (s.size() > 1) ? ((ll)(s[0] - '0')*calc(s.substr(1))) : 0
    });
}
int main() {
    string s;
    cin >> s;
    cout << calc(s) << endl;
}

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
inputCopy
390
outputCopy
216
inputCopy
7
outputCopy
7
inputCopy
1000000000
outputCopy
387420489
Note
In the first example the maximum product is achieved for 389 (the product of digits is 3⋅8⋅9=216).

In the second example the maximum product is achieved for 7 (the product of digits is 7).

In the third example the maximum product is achieved for 999999999 (the product of digits is 99=387420489).

标签：digits,1143,product,q1,ll,maximum,Nirvana,include
来源： https://blog.csdn.net/weixin_43870649/article/details/88981238