B. Vanya and Books( Codeforces Round #308 (Div. 2) 简单题)

B. Vanya and Books

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard 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.

Sample test(s)

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：为1-n之中有多少个数字。比如10有两个数字。100有三个数字。

点击打开链接

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

using namespace std;

__int64 sum;
__int64 a[] = {0,9,90,900,9000,90000,900000,9000000,90000000,900000000};
__int64 b[] = {0,10,100,1000,10000,100000,1000000,10000000,100000000,1000000000,10000000000};
__int64 n;

int main(){
    while(scanf("%I64d",&n)!=EOF){
        sum = 0;
        for(int i=1;i<=10;i++){
            if(n>=b[i]){
                sum += a[i]*i;
            }else{
                sum += (n - b[i-1] + 1)*i;
                break;
            }
        }
        if(n<10)
        {
            sum--;
        }
        printf("%I64d
",sum);
    }
}