题干：
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
) 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
, i.e. 7.
- The strength of second ranger remains the same, i.e. 7.
- The strength of third ranger is updated to
, i.e. 11.
- The strength of fourth ranger remains the same, i.e. 11.
- The strength of fifth ranger is updated to
, 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 7Input
2 100000 569
605 986
Output
986 605题目大意：
给你一组数，进行k组操作，每组操作中包含两个步骤，首先排序，然后对下标为奇数的数字进行异或操作。最后问你进行k组操作之后的数组中的最大值和最小值。
解题报告：
不是我就想知道这题打死也想不到会出现循环节的情况吧？？！！？这也太、、可能这是异或运算的一种特性？反正我是第一次遇到，题目中会猜他存在循环节然后进行暴力的。。。然后，循环节是4不是2、所以其实代码中的(ci-3)是不需要进行判断的。（所以对于这种不知道循环节是几的这种题，保险起见建议那种精彩代码，直接Node结构体记录曾经算过的状态，并且用vector存下来！！技巧啊）
AC代码：
#include<bits/stdc++.h>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;
int a[MAX];
int mx[MAX],mi[MAX];
int main()
{int n,k,x,ci;cin>>n>>k>>x;mx[0]=0,mi[0]=100005;for(int i = 1; i<=n; i++) scanf("%d",a+i),mx[0] = max(mx[0],a[i]),mi[0] = min(mi[0],a[i]);for(ci = 1; ci<=k; ci++) {sort(a+1,a+n+1);mx[ci] = -100005;//a[n];是错的 mi[ci] = 100005;//a[1];是错的 for(int i = 1; i<=n; i++) {if(i&1) a[i] = a[i] ^ x;mi[ci] = min(a[i],mi[ci]);mx[ci] = max(a[i],mx[ci]);}if(ci > 4) {if(mi[ci] == mi[ci-1] && mi[ci] == mi[ci-2] && mi[ci] == mi[ci-3] && mi[ci] == mi[ci-4])if(mx[ci] == mx[ci-1] && mx[ci] == mx[ci-2] && mx[ci] == mx[ci-3] && mx[ci] == mx[ci-4]) {printf("%d %d\n",mx[ci],mi[ci]);return 0 ;}}}
// sort(a+1,a+n+1);printf("%d %d\n",mx[ci-1],mi[ci-1]);return 0 ;}1WA代码：
需要i++,而不是i+=2直接做，因为你想啊，你这一轮操作的mx和mi，都是所有数字的，而不是选出来那些数的啊！！另外，不能直接每次都mx[ci] = a[n]这样，因为这一轮操作之后可能就没有这个最大值了，可能就被覆盖成别的值了。，。。所以说啊，写代码的时候一定要注意是否代码中存在小问题，这些问题叠加到一块就很致命了！！
#include<bits/stdc++.h>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;
int a[MAX];
int mx[MAX],mi[MAX];
int main()
{int n,k,x,ci;cin>>n>>k>>x;mx[0]=0,mi[0]=100005;for(int i = 1; i<=n; i++) scanf("%d",a+i),mx[0] = max(mx[0],a[i]),mi[0] = min(mi[0],a[i]);for(ci = 1; ci<=k; ci++) {sort(a+1,a+n+1);mx[ci] = -100005;//a[n];是错的 mi[ci] = 100005;//a[1];是错的 for(int i = 1; i<=n; i+=2) {//这样写应该也是错的 a[i] = a[i] ^ x;mi[ci] = min(a[i],mi[ci]);mx[ci] = max(a[i],mx[ci]);}if(ci > 4) {if(mi[ci] == mi[ci-1] && mi[ci] == mi[ci-2] && mi[ci] == mi[ci-3] && mi[ci] == mi[ci-4])if(mx[ci] == mx[ci-1] && mx[ci] == mx[ci-2] && mx[ci] == mx[ci-3] && mx[ci] == mx[ci-4]) {printf("%d %d\n",mx[ci],mi[ci]);return 0 ;}}}sort(a+1,a+n+1);printf("%d %d\n",a[n],a[1]);return 0 ;}
另一个精彩的代码：
把每次变化的都存起来。然后得到一个新的状态的时候，先去跟之前存起来的那些状态比较，如果发现又相同的状态，那么就表示找到了循环节了。
之后求出来循环节的长度，然后取模，得到最后的答案。
（代码中可以直接return 0的，不需要goto，因为是codeforce的题。、。。）
#include <bits/stdc++.h>using namespace std;
const int MAXN=1e5+7;
int n,k,x;
int cnt;
struct node { //用来存储已经出现过的状态int num[MAXN];int MAX;int MIN;node() {MAX=0;MIN=1e9;}bool operator ==(const node &a)const {for(int i=0; i<n; ++i) {if(a.num[i]!=num[i])return 0;}return 1;}
} p;
vector<node>q;
int check() {for(int i=0; i<cnt; ++i) {if(q[i]==p)return i;}return -1;
}
int main() {int i;
xx:while(~scanf("%d%d%d",&n,&k,&x)) {q.clear();p.MAX=0;p.MIN=1e9;for(i=0; i<n; ++i) {scanf("%d",&p.num[i]);p.MAX=max(p.MAX,p.num[i]);p.MIN=min(p.MIN,p.num[i]);}sort(p.num,p.num+n);q.push_back(p);cnt=0;int pos;while(cnt<k) {cnt++;p.MAX=0;p.MIN=1e9;for(i=0; i<n; i+=2) {p.num[i]^=x;}for(i=0; i<n; ++i) {p.MAX=max(p.MAX,p.num[i]);p.MIN=min(p.MIN,p.num[i]);}sort(p.num,p.num+n);pos=check();if(pos!=-1) { //找到循环节了int t=cnt-pos;//循环节长度k=(k-pos)%t+pos;//取模之后的答案的下标printf("%d %d\n",q[k].MAX,q[k].MIN);goto xx;//找到了就直接重新读取下一组}q.push_back(p);}printf("%d %d\n",p.MAX,p.MIN);//表示没有找到循环节}return 0;
}
