【C++】Digit sum 2

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本文介绍了一个编程问题，如何在C++中找到一个不超过给定整数N的正整数的最大十进制数码和。通过编写一个sum函数计算单个数字之和，然后使用while循环和条件判断找到最优解。

题目描述

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.

输入

Input is given from Standard Input in the following format:

N

输出

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

样例输入 Copy

样例输出 Copy

提示

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

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
int sum(LL x)
{
	int ans = 0;
	long long temp = x;
	while (x)
	{
		ans += x % 10;
		x /= 10;
	}
	return ans;
}
int main()
{
	LL n;
	cin >> n;
	int s = sum(n);
	LL temp = n, w = 1;
	while (temp >= 10)
	{
		temp /= 10;
		w *= 10;
	}
	cout << max(s, sum(n / w * w - 1)) << '\n';
	return 0;
}