AtCoder-3867

Find the maximum possible sum of the digits (in base 10) of a positive integer not greater than N.

Constraints

• 1≤N≤10¹⁶
• N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the maximum possible sum of the digits (in base 10) of a positive integer not greater than N.

Sample Input 1

100

Sample Output 1

18

For example, the sum of the digits in 99 is 18, which turns out to be the maximum value.

Sample Input 2

9995

Sample Output 2

35

For example, the sum of the digits in 9989 is 35, which turns out to be the maximum value.

Sample Input 3

3141592653589793

Sample Output 3

137

题解：

其实只要比较原数所有位上的和与原数最大为减一其他为变为9后所有的和就可以啦。

AC代码为：

#include
#include
#include
#include
using namespace std;

int Sum(long long n) 
{
	int res = 0;
	while (n > 0) 
	{
		res += n % 10;
		n /= 10;
	}
	return res;
}
int main() 
{
	long long n, temp;
	cin >> n;
	int ans = Sum(n);
	temp = n;
	long long cnt = 1;
	while (temp >= 10) 
	{
		temp /= 10;
		cnt *= 10;
	}
	ans = max(ans, Sum(n / cnt * cnt - 1));
	cout << ans << endl;
	return 0;
}

转载于:https://www.cnblogs.com/songorz/p/9386610.html