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

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.

題意 ：給你一個整數n，問你從1到n一共有多少位。比如n = 13，1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13一共17位。

f[i]表示從1到 10i 一共有多少位，以753為例，從100到753都是3位數，是以答案就是f[2]+653*3。

轉載請注明出處：尋找&星空の孩子 

題目連結：http://codeforces.com/contest/552/problem/B

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include<string>
#include<map>
#define LL __int64
using 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;
}
           

或者

#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;
}