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  • Codeforces 932 B.Recursive Queries-前缀和 (ICM Technex 2018 and Codeforces Round #463 (Div. 1 + Div. 2, combined))

    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

    这道题就是直接暴力应该会超时,用前缀和处理一下就可以了(吐槽,一开始都没看懂题,本咸鱼就没读懂过带公式的题(╥╯^╰╥))

    代码:

    //B. Recursive Queries-前缀和
    #include<iostream>
    #include<cstring>
    #include<cstdio>
    #include<cmath>
    #include<algorithm>
    #include<queue>
    #include<stdlib.h>
    using namespace std;
    const int maxn=1e6+10;
    int a[10][maxn];
    int fun(int x){
        if(x<10)return x;
        int sum=1;
        while(x){
            sum*=x%10>0?x%10:1;
            x/=10;
        }
        return fun(sum);
    }
    void qianzhuihe(){
        for(int i=1;i<=1000000;i++)
           a[fun(i)][i]++;
        for(int i=1;i<10;i++){
            for(int j=1;j<=1000000;j++)
                a[i][j]+=a[i][j-1];
        }
    }
    int main(){
        qianzhuihe();
        int t;
        scanf("%d",&t);
        while(t--){
            int l,r,k;
            scanf("%d%d%d",&l,&r,&k);
            printf("%d
    ",a[k][r]-a[k][l-1]);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/ZERO-/p/9711342.html
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