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  • Codeforces Round #272 (Div. 2) D. Dreamoon and Sets 构造

    D. Dreamoon and Sets

    题目连接:

    http://www.codeforces.com/contest/476/problem/D

    Description

    Dreamoon likes to play with sets, integers and . is defined as the largest positive integer that divides both a and b.

    Let S be a set of exactly four distinct integers greater than 0. Define S to be of rank k if and only if for all pairs of distinct elements si, sj from S, .

    Given k and n, Dreamoon wants to make up n sets of rank k using integers from 1 to m such that no integer is used in two different sets (of course you can leave some integers without use). Calculate the minimum m that makes it possible and print one possible solution.

    Input

    The single line of the input contains two space separated integers n, k (1 ≤ n ≤ 10 000, 1 ≤ k ≤ 100).

    Output

    On the first line print a single integer — the minimal possible m.

    On each of the next n lines print four space separated integers representing the i-th set.

    Neither the order of the sets nor the order of integers within a set is important. If there are multiple possible solutions with minimal m, print any one of them.

    Sample Input

    1 1

    Sample Output

    5
    1 2 3 5

    Hint

    题意

    让你构造n个集合,使得每个集合里面都有四个元素,且两两元素之间的gcd都是k

    让你使得其中最大数最小

    题解:

    其实就是样例的构造方法,这个得自己手玩一玩才行

    代码

    #include<bits/stdc++.h>
    using namespace std;
    
    int main()
    {
        int n,k;
        scanf("%d%d",&n,&k);
        printf("%d
    ",(n-1)*k*6+5*k);
        for(int i=0;i<n;i++)
            printf("%d %d %d %d
    ",i*k*6+k,i*k*6+2*k,i*k*6+3*k,i*k*6+5*k);
    }
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  • 原文地址:https://www.cnblogs.com/qscqesze/p/5794679.html
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