Codeforces Round #549 div2 1143-B Nirvana 题解

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

一开始看到这道题的时候还是有点伤脑筋的，一直没想出来，想了十多分钟后想到办法后AC
代码使用字符串模拟数字求得结果，getvalue()函数是求得结果，calc()函数是递归函数，用于求得最大值
这个采用的递归搜索，设定结果函数为calc()，字符串长度为n，则calc返回以下其中一个最大解：

1. getvalue( {s[0], s[1], ..., s[n-1]} )
2. getvalue( {s[0]-1, 9, ..., 9} ) //连续n-1个9
3. getvalue( {9, 9, ..., 9} ) //连续n-1个9
4. s[0] * calc( {s[1], s[2], ..., s[n-1]} )

代码如下：

#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;
}