Codeforces768C Jon Snow and his Favourite Number

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最新推荐文章于 2022-01-29 00:43:53 发布

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分类专栏： Codeforces ----枚举

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在一场与异鬼的战斗中，Jon Snow希望通过对他手下的游骑兵的力量进行特定操作来提高他们的战斗力。具体来说，他会将游骑兵按力量升序排列，并将其与他的幸运数字进行异或运算，以期获得更强的战士。

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Jon Snow now has to fight with White Walkers. He has n rangers, each of which has his own strength. Also Jon Snow has his favourite number x. Each ranger can fight with a white walker only if the strength of the white walker equals his strength. He however thinks that his rangers are weak and need to improve. Jon now thinks that if he takes the bitwise XOR of strengths of some of rangers with his favourite number x, he might get soldiers of high strength. So, he decided to do the following operation k times:

Arrange all the rangers in a straight line in the order of increasing strengths.
Take the bitwise XOR (is written as $\text{[math]}$ ) of the strength of each alternate ranger with x and update it's strength.

Suppose, Jon has 5 rangers with strengths [9, 7, 11, 15, 5] and he performs the operation 1 time with x = 2. He first arranges them in the order of their strengths, [5, 7, 9, 11, 15]. Then he does the following:

The strength of first ranger is updated to $\text{[math]}$ , i.e. 7.
The strength of second ranger remains the same, i.e. 7.
The strength of third ranger is updated to $\text{[math]}$ , i.e. 11.
The strength of fourth ranger remains the same, i.e. 11.
The strength of fifth ranger is updated to $\text{[math]}$ , i.e. 13.

The new strengths of the 5 rangers are [7, 7, 11, 11, 13]

Now, Jon wants to know the maximum and minimum strength of the rangers after performing the above operations k times. He wants your help for this task. Can you help him?

Input

First line consists of three integers n, k, x (1 ≤ n ≤ 105, 0 ≤ k ≤ 105, 0 ≤ x ≤ 103) — number of rangers Jon has, the number of times Jon will carry out the operation and Jon's favourite number respectively.

Second line consists of n integers representing the strengths of the rangers a1, a2, ..., an (0 ≤ ai ≤ 103).

Output

Output two integers, the maximum and the minimum strength of the rangers after performing the operation k times.

Examples

input
5 1 2
9 7 11 15 5

output
13 7

input
2 100000 569
605 986

output
986 605


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题目的意思是把一组数从小到大排列，奇数位与给定的数异或，重复m次，问最后最大的数和最小的数




可以找循环节，因为数字小也可以纯暴力跑。







#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <string>
#include <set>
#include <map>
using namespace std;
int cnt[2][2005];
int n,k,m;



int main()
{
    int x;
    while(~scanf("%d%d%d",&n,&m,&k))
    {
        memset(cnt,0,sizeof(cnt));

        for(int i=0; i<n; i++)
        {
            scanf("%d",&x);
            cnt[0][x]++;
        }
        for(int i=0; i<m; i++)
        {
            bool flag=0;
            for(int j=0; j<2000; j++)
            {
                if(cnt[0][j]>0||cnt[1][j]>0)
                {
                    if(cnt[0][j]%2==0)
                    {
                        int s=j^k;
                        if(s<=j)
                            cnt[0][s]+=cnt[0][j]/2;
                        else
                            cnt[1][s]+=cnt[0][j]/2;
                        cnt[0][j]/=2;
                        cnt[0][j]+=cnt[1][j];
                        cnt[1][j]=0;
                    }
                    else
                    {
                        int s=j^k;
                        if(flag==0)
                        {
                            if(s<=j)
                                cnt[0][s]+=cnt[0][j]/2+1;
                            else
                                cnt[1][s]+=cnt[0][j]/2+1;
                            cnt[0][j]/=2;
                            cnt[0][j]+=cnt[1][j];
                            cnt[1][j]=0;
                            flag=1;
                        }
                        else
                        {
                            if(s<=j)
                                cnt[0][s]+=cnt[0][j]/2;
                            else
                                cnt[1][s]+=cnt[0][j]/2;
                            cnt[0][j]/=2;
                            cnt[0][j]++;
                            cnt[0][j]+=cnt[1][j];
                            cnt[1][j]=0;
                            flag=0;
                        }

                    }
                }
            }
        }
        int mn,mx;
        for(int i=0;i<=2000;i++)
            {
                if(cnt[0][i]>0)
                {
                    mn=i;
                    break;
                }
            }
             for(int i=2000;i>=0;i--)
            {
                if(cnt[0][i]>0)
                {
                    mx=i;
                    break;
                }
            }
        printf("%d %d\n",mx,mn);
    }
    return 0;
}