hdu5344 MZL's xor(水题)

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MZL's xor

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 488    Accepted Submission(s): 342

Problem Description

MZL loves xor very much.Now he gets an array A.The length of A is n.He wants to know the xor of all (Ai+Aj)(1≤i,j≤n)
The xor of an array B is defined as B1 xor B2...xor Bn

 

Input

Multiple test cases, the first line contains an integer T(no more than 20), indicating the number of cases.
Each test case contains four integers:n,m,z,l
A1=0,Ai=(Ai−1∗m+z) mod l
1≤m,z,l≤5∗105,n=5∗105

 

Output

For every test.print the answer.

 

Sample Input

2
3 5 5 7
6 8 8 9

 

Sample Output

14 16

两个相同的数异或之后等于0，最终只剩下Ai+Ai的异或值

 /**
  * code generated by JHelper
  * More info: https://github.com/AlexeyDmitriev/JHelper
  * @author xyiyy @https://github.com/xyiyy
  */

 #include <iostream>
 #include <fstream>

 //#####################
 //Author:fraud
 //Blog: http://www.cnblogs.com/fraud/
 //#####################
 //#pragma comment(linker, "/STACK:102400000,102400000")
 #include <iostream>
 #include <sstream>
 #include <ios>
 #include <iomanip>
 #include <functional>
 #include <algorithm>
 #include <vector>
 #include <string>
 #include <list>
 #include <queue>
 #include <deque>
 #include <stack>
 #include <set>
 #include <map>
 #include <cstdio>
 #include <cstdlib>
 #include <cmath>
 #include <cstring>
 #include <climits>
 #include <cctype>

 using namespace std;
 #define rep(X, N) for(int X=0;X<N;X++)
 typedef long long ll;

 class hdu5344 {
 public:
     void solve(std::istream &in, std::ostream &out) {
         int t;
         in >> t;
         while (t--) {
             ll m, z, l, n;
             in >> n >> m >> z >> l;
             ll x = ;
             ll ans = ;
             rep(i, n) {
                 ans ^= (x + x);
                 x = (x * m + z) % l;
             }
             out << ans << endl;
         }
     }
 };

 int main() {
     std::ios::sync_with_stdio(false);
     std::cin.tie();
     hdu5344 solver;
     std::istream &in(std::cin);
     std::ostream &out(std::cout);
     solver.solve(in, out);
     return ;
 }