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  • Vanya and Books (数位DP)

    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.

    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
    ", sum); 
        }
        return 0;
    }

     

     

     

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  • 原文地址:https://www.cnblogs.com/soTired/p/5418340.html
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