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  • C. Book Reading (Codeforces Round #582 (Div. 3))

    Polycarp is reading a book consisting of nn pages numbered from 11 to nn. Every time he finishes the page with the number divisible by mm, he writes down the last digit of this page number. For example, if n=15n=15 and m=5m=5, pages divisible by mm are 5,10,155,10,15. Their last digits are 5,0,55,0,5 correspondingly, their sum is 1010.

    Your task is to calculate the sum of all digits Polycarp has written down.

    You have to answer qq independent queries.

    Input

    The first line of the input contains one integer qq (1q10001≤q≤1000) — the number of queries.

    The following qq lines contain queries, one per line. Each query is given as two integers nn and mm (1n,m10161≤n,m≤1016) — the number of pages in the book and required divisor, respectively.

    Output

    For each query print the answer for it — the sum of digits written down by Polycarp.

    Example
    input
    Copy
    7
    1 1
    10 1
    100 3
    1024 14
    998244353 1337
    123 144
    1234312817382646 13
    
    output
    Copy
    1
    45
    153
    294
    3359835
    0
    427262129093995
    

     


    分析:计算出n范围内m的个位数的倍数的个位数的和


    #include <bits/stdc++.h>
     
    #define TOP 10001
     
    using namespace std;
     
    typedef long long ll;
     
    int main()
    {
        ios_base::sync_with_stdio(0);
        cin.tie(0);
     
        map<int, vector<int> > r;
        int q;
        ll m, n, coc, rest, fin, ans;
        vector<int> aux;
     
        r[1] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 0};
        r[2] = {2, 4, 6, 8, 0};
        r[3] = {3, 6, 9, 2, 5, 8, 1, 4, 7, 0};
        r[4] = {4, 8, 2, 6, 0};
        r[5] = {5, 0};
        r[6] = {6, 2, 8, 4, 0};
        r[7] = {7, 4, 1, 8, 5, 2, 9, 6, 3, 0};
        r[8] = {8, 6, 4, 2, 0};
        r[9] = {9, 8, 7, 6, 5, 4, 3, 2, 1 ,0};
     
     
        cin >> q;
        while(q--){
            cin >> n >> m;
            coc = n / m;
            fin = m % 10;
            ans = 0;
            if(fin != 0){
                aux = r[fin];
                n = aux.size();
                rest = coc % n;
                coc /= n;
                for(int i = 0; i < n; ++i)
                    ans += aux[i] * (coc + (i < rest));
            }
            cout << ans << '
    ';
        }
        return 0;
    }

     


    所遇皆星河
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  • 原文地址:https://www.cnblogs.com/Shallow-dream/p/11517777.html
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