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  • Codeforces Round #463 (Div. 1 + Div. 2, combined) B. Recursive Queries (打表,前缀和)

    B. Recursive Queries

    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Let us define two functions f and g on positive integer numbers.

    You need to process Q queries. In each query, you will be given three integers lr and k. You need to print the number of integers xbetween l and r inclusive, such that g(x) = k.

    Input

    The first line of the input contains an integer Q (1 ≤ Q ≤ 2 × 105) representing the number of queries.

    Q lines follow, each of which contains 3 integers lr and k (1 ≤ l ≤ r ≤ 106, 1 ≤ k ≤ 9).

    Output

    For each query, print a single line containing the answer for that query.

    Examples
    input
    Copy
    4
    22 73 9
    45 64 6
    47 55 7
    2 62 4
    output
    1
    4
    0
    8
    input
    Copy
    4
    82 94 6
    56 67 4
    28 59 9
    39 74 4
    output
    3
    1
    1
    5
    Note

    In the first example:

    • g(33) = 9 as g(33) = g(3 × 3) = g(9) = 9
    • g(47) = g(48) = g(60) = g(61) = 6
    • There are no such integers between 47 and 55.
    • g(4) = g(14) = g(22) = g(27) = g(39) = g(40) = g(41) = g(58) = 4

    题意主要就是,g(x)等于各个数位和的乘积(不包含0),给出a,b,c,问ab之间有多少个g[x]=c。这个题目第一反应就是打表,做一个映射,然后再求前缀和。注意数组开大一点,不然会RE。还有一个坑点就是减的时候注意是开区间,所以后一个前缀和是a-1。

    #include<bits/stdc++.h>
    
    using namespace std;
    #define maxn 2000005
    int n;
    int num[maxn];
    int cnt[maxn][10];
    
    int deal(int n)
    {
        int sum=1;
        while(n)
        {
            if(n%10!=0)
                sum*=(n%10);
            n/=10;
        }
        return sum;
    }
    
    void create()
    {
        for(int i=1; i<=maxn; i++)
        {
            int t=deal(i);
            while(t>=10)
                t=deal(t);
            num[i]=t;
            cnt[i][t]++;
        }
    }
    
    
    int main()
    {
        create();
        for(int i=1; i<=9; i++)
            for(int j=2; j<=maxn; j++)
                cnt[j][i]+=cnt[j-1][i];
        int x,y;
        /*while(cin>>x>>y)
                    cout<<cnt[x][y]<<endl;*/
        scanf("%d",&n);
        for(int i=0; i<n; i++)
        {
            int a,b,c;
            scanf("%d %d %d",&a,&b,&c);
            int ans=cnt[b][c]-cnt[a-1][c];    //开区间注意a-1
            printf("%d
    ",ans);
        }
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/youchandaisuki/p/8728323.html
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