Codeforces 552B - Vanya and Books(数学)

B. Vanya and Books

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

Vanya got an important task — he should enumerate books in the library and label each book with its number. Each of the n books should be assigned with a number from 1 to n. Naturally, distinct books should be assigned distinct numbers.

Vanya wants to know how many digits he will have to write down as he labels the books.

Input

The first line contains integer n (1 ≤ n ≤ 109) — the number of books in the library.

Output

Print the number of digits needed to number all the books.

Examples

input

13

output

17

input

4

output

4

Note

Note to the first test. The books get numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, which totals to 17 digits.

Note to the second sample. The books get numbers 1, 2, 3, 4, which totals to 4 digits.

题意:

给出一个n,输出从1-n的所有数,一共要输出几个数字.

(比如17这个数要输出2个数字).

解题思路:

1-9有9个数

10-99有90个数

100-999有900个数

那么对于10^digit - 10(digit+1)-1 一共有9*10^digit个数.

对于最大长度的数出现的次数为 (n - pow(10, dight-1) + 1) * dight

剩下的累加计算即可.

AC代码:

#include <bits/stdc++.h>
using namespace std;
#define int long long
int pow(int a, int b)
{
int tmp = a;
for(int i = 1; i <= b; i++)
a *= tmp;
return a;
}
main()
{
int n;
cin >> n;
int tmp = n;
int dight = 0;
while(tmp)
dight++, tmp /= 10;
int sum = 0;
for(int i = 1;i <= dight-1; i++)
sum += 9 * pow(10, i-1) * i;
sum += (n - pow(10, dight-1) + 1) * dight;
cout << sum << endl;
return 0;
}