Vanya and Books (数位DP)-优快云博客

本文针对一个具体的图书编号问题，提供了一种高效的解决方案。通过分析图书数量与所需数字总数之间的关系，设计了一个能够快速计算出所有图书编号所需数字总数的算法。文章通过一个具体的示例详细解释了算法的工作原理。

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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.
脑洞一波, 强行打表:
 
 
    #include <cstdio>
typedef long long LL;
LL a[100]={0, 9, 180, 2700, 36000, 450000, 5400000, 63000000, 720000000, 8100000000, 90000000000};
LL count(LL a, LL b)
{
    if(b==0)
        return 1;
    LL sum=1;
    for(int i=0; i<b; i++)
        sum*= a;
    return sum;
}
int main()
{
    LL n; 
    while(scanf("%lld", &n) != EOF)
    {
        LL sum =0;
        int cnt =0;
        LL O=n;
        while(n)
        {
            if(n <= 9) break;
            LL Q=n;
            if(Q/10 != 0)
         　　{
                sum +=a[++cnt];
                n/=10;
                continue;
            }
        }
        LL dig= cnt+1;
        sum= sum+(O-count(10, cnt)+1)*dig;
        printf("%lld\n", sum); 
    }
    return 0;
} 
   
 
 
 
 