Simple game (easygame)

$$Simplesingletripplay\left(EASYGAmE\right)$$

Topic link: $SSLratiorace.\; 1.\; 50.\; 7$

topic

One day, little R was going to find little h to go swimming. When he found little h, he found that little h was writing a column of numbers in pain, $1$ ， $2$ ， $3$ ， $\dots$ ， $n$ , so I asked little h the reason for his pain, and little h told him that now he has to count$.\; 1..N-$ these numbers inside,11How many times does$1$ appear, such as$n=1\; At1$ , there are$1,10,11$ appeared$4$ times$1$ , now$n,$ can you give the answer quickly?

enter

One line, an integer $n$

Output

An integer representing $.\; 1..N-$ in$1$ number of occurrences.

Sample input

11

Sample output

4

data range

For $.\; 30\%$ Data:$n<=1000;$
for$.\; 100\%$ Data:$n<=maxlongint;$

Ideas

This question is a mathematics question and should be discussed separately.

We are asking
for one by one: For the ones, 1 1 appears here for every ten$1$ Ge$1$ ; For ten digits,10 10appears here for every hundred$10$ Ge$1$ ; For hundreds of places, there will be100 100if there are no thousands$100$ Ge$1$ 。

The number in front of it is the number of each, or ten, hundred, etc.
(Think $233$ , for the tens place, there is$2$ Ge$10$ ）

Then look at this bit, you can find that if this bit is $0$ , it is normal. If this bit is greater than$1$ , there will be one more each, or ten, one hundred, etc.

But if this bit is equal to $1$ ?
Then we found that the number behind it is how many more times$+1$ .
Why add one? Because for example$10$ , when you see the tens place, although the number behind is$0$ , but it can be seen that1 1appears in the ten digits$1$ time$1$ . Because$.\; 10$ which is a ten$1$ of.

Code

#include<cstdio>
#define ll long long

using namespace std;

ll n, ans;
ll more, times, sheng;

int main() {
    
    
	scanf("%lld", &n);//读入
	
	for (ll wei = 1; wei <= n; wei *= 10) {
    
    //枚举每一位
		more = n % (wei * 10) / wei;//这一位
		times = n / (wei * 10);//它前面组成的数字
		sheng = n % wei;//它后面组成的数字
		if (more > 1) ans += times * wei + wei;
			else if (more == 1) ans += times * wei + sheng + 1;//分类讨论
				else ans += times * wei;
	}
	
	printf("%lld", ans);//输出
	
	return 0;
}