【codeforces 768C】Jon Snow and his Favourite Number

【题目链接】:http://codeforces.com/contest/768/problem/C

【题意】

给你n个数字;
让你每次把这n个数字排序;
然后对奇数位的数字进行异或操作,然后对新生成的序列再重复上述操作总共k次;
问最后的序列的最小值和最大值分别是多少;

【题解】

可以用计数排序来搞;
因为数字最大为1000;
所以将近O(K*1000)的复杂度;
完全可以的;
每次从小到大(0到1000(枚举数字;
然后根据该数字的个数判断该数字有几个数字要进行异或,几个数字保持原样.
然后在新的计数排序数组里面记录一下就好;
最后一遍循环搞出最大最小值就好;

【完整代码】

#include <bits/stdc++.h>using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define LL long long#define rep1(i,a,b) for (int i = a;i <= b;i++)#define rep2(i,a,b) for (int i = a;i >= b;i--)#define mp make_pair#define pb push_back#define fi first#define se second#define rei(x) scanf("%d",&x)#define rel(x) scanf("%lld",&x)typedef pair<int,int> pii;typedef pair<LL,LL> pll;const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};const double pi = acos(-1.0);const int N = 1e3+500;int n,k,xx;int bo[2][N];int main(){    //freopen("F:\\rush.txt","r",stdin);    rei(n),rei(k),rei(xx);    rep1(i,1,n)    {        int y;        rei(y);        bo[0][y]++;    }    int now = 0;    rep1(i,1,k)    {        rep1(j,0,1100)            bo[1-now][j] = 0;        int x = 0;        rep1(j,0,1100)            if (bo[now][j]>0)            {                int ju = bo[now][j]&1;                if (x==0)                {                    if (ju)                    {                        bo[1-now][j^xx]+=((bo[now][j]+1)/2);                        bo[1-now][j]+=(bo[now][j]-((bo[now][j]+1)/2));                        x = 1;                    }                    else                    {                        bo[1-now][j^xx]+=bo[now][j]/2;                        bo[1-now][j]+=(bo[now][j]-(bo[now][j]/2));                        x = 0;                    }                }                else                {                    if (ju)                    {                        bo[1-now][j^xx]+=((bo[now][j]-1)/2);                        bo[1-now][j]+=(bo[now][j]-((bo[now][j]-1)/2));                        x = 0;                    }                    else                    {                        bo[1-now][j^xx]+=((bo[now][j])/2);                        bo[1-now][j]+=(bo[now][j]-((bo[now][j])/2));                        x = 1;                    }                }            }        now = 1-now;    }    rep2(i,1100,0)        if (bo[now][i]>0)        {            printf("%d ",i);            break;        }    rep1(i,0,1100)        if (bo[now][i]>0)        {            printf("%d\n",i);            break;        }    return 0;}