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  • D-D. Gluttony(构造)

    D. Gluttony
    time limit per test2 seconds
    memory limit per test256 megabytes
    inputstandard input
    outputstandard output
    You are given an array a with n distinct integers. Construct an array b by permuting a such that for every non-empty subset of indices S = {x1, x2, …, xk} (1 ≤ xi ≤ n, 0 < k < n) the sums of elements on that positions in a and b are different, i. e.

    Input
    The first line contains one integer n (1 ≤ n ≤ 22) — the size of the array.

    The second line contains n space-separated distinct integers a1, a2, …, an (0 ≤ ai ≤ 109) — the elements of the array.

    Output
    If there is no such array b, print -1.

    Otherwise in the only line print n space-separated integers b1, b2, …, bn. Note that b must be a permutation of a.

    If there are multiple answers, print any of them.

    Examples
    inputCopy
    2
    1 2
    outputCopy
    2 1
    inputCopy
    4
    1000 100 10 1
    outputCopy
    100 1 1000 10
    Note
    An array x is a permutation of y, if we can shuffle elements of y such that it will coincide with x.

    Note that the empty subset and the subset containing all indices are not counted.

    题意:要求每段区间的数字和都不相同的排列。
    思路:需要有一个满足题意排列策略,一个可行的办法是将每个数字映射为它排序后位置的后一位(最后一位映射为第一位)。证明:每次选到一个区间,num[i] ~ num[i + k],排列后得到 num[i + x](排序后下一位,最大则映射为最小) + num[i + 1 + y]…没有选到最大的数时,两者做差可得结果大于0。选到最大的数时,假设全部选上(实际不可能全部选上),作差得0,去掉任意一个原数中非最大值的数和映射后的结果,作差结果必然小于0,(可以自己手算模拟一下)。结论:区间映射后两者作差不为0,故不相等。

    #include <bits/stdc++.h>
    
    using namespace std;
    
    struct node
    {
        int x;
        int id;
    }num[30];
    
    int cmp(node a,node b)
    {
        return a.x < b.x;
    }
    
    int main()
    {
        int n,_hash[30];
        scanf("%d",&n);
        for(int i = 1;i <= n;i ++)
        {
            scanf("%d",&num[i].x);
            num[i].id = i;
        }
        sort(num + 1,num + 1 + n,cmp);
        for(int i = 1;i <= n;i++)
        {
            _hash[num[i].id] = i;
        }
        for(int i  =1;i <= n;i++)
        {
            int t = (_hash[i] + 1) % n;
            if(t == 0)
                t = n;
            printf("%d ",num[t].x);
        }
        return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/tomjobs/p/10617580.html
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