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

B. Vanya and Books (Codeforces Round #308 (Div. 2 )),
B. Vanya and Bookstime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

Vanya got an important task-he shocould enumerate books in the library and label each book with its number. Each ofNBooks shoshould be assigned with a number from 1N. Naturally, distinct books shoshould 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 integerN(1 digit ≤ DigitNBandwidth ≤ limit 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, clerk 2, clerk 3, clerk 4, clerk 5, clerk 6, clerk 7, clerk 8, clerk 9, clerk 10, clerk 11, clerk 12, listen 13, which totals to 17 digits.

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

Question: Give you an integer n and ask how many digits are there from 1 to n. For example, n = 13, 1, listen 2, listen 3, listen 4, Listen 5, listen 6, listen 7, listen 8, then 9, then 10, then 11, then 12, listen 13 has 17 digits in total.

F [I] indicates from 1 10i The total number of digits. For example, from 753 to 100 is three digits, so the answer is f [2] + 753*3.

Reprinted, please specify the Source: search for & STAR kids

Link: http://codeforces.com/contest/552/problem/ B

#include <stdio.h>#include <string.h>#include <algorithm>#include<string>#include<map>#define LL __int64using namespace std;const int MAXN = 1005;LL n;LL ans = 0;LL a[12]= {0,9,99,999,9999,99999,999999,9999999,99999999,999999999,9999999999};int main(){    while(~scanf("%I64d",&n))    {        ans = 0;        LL i,j,k = 1;        if(n<10)        {            printf("%I64d\n",n);            return 0;        }        for(i = 1; i<=10; i++)        {            k *= 10;            if(n<k)            {                ans+=(n-k/10+1)*i;              //   printf("%I64d %I64d %I64d\n",i,ans,(n-k/10+1)*i);                break;            }           /* else if(n==k)            {                ans+=(i+1);                break;            }*/            else            {                ans+=(a[i]-a[i-1])*i;            }            // printf("%d %I64d\n",i,ans);        }        printf("%I64d\n",ans);    }    return 0;}

Or

#include <bits/stdc++.h>using namespace std;int n;long long f[12];long long t[12];int main() {    cin >> n;    t[1] = 10;    for (int i = 2; i <= 10; i++)        t[i] = t[i - 1] * 10;    f[1] = 11;    for (int i = 2; i <= 10; i++)        f[i] = f[i - 1] + (t[i] - t[i - 1]) * i + 1;    long long ans;    long long base;    for (int i = 0; i <= n; i++) {        if (t[i] > n) {            base = i - 1;            break;        }    }    ans = f[base] + (n - t[base]) * (base + 1);    cout << ans << endl;    return 0;}