hdu 5344 MZL's xor

MZL's xor
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 310    Accepted Submission(s): 225
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
   
   
   
   

 
标题:すべてのa[i]+a[j]の異或値を求めて、a[i]+a[j]とa[j]+a[i]が異或落するためa[i]+a[i]だけを考慮します
コード:

#include <stdio.h>  
#include <ctime>  
#include <math.h>  
#include <limits.h>  
#include <complex>  
#include <string>  
#include <functional>  
#include <iterator>  
#include <algorithm>  
#include <vector>  
#include <stack>  
#include <queue>  
#include <set>  
#include <map>  
#include <list>  
#include <bitset>  
#include <sstream>  
#include <iomanip>  
#include <fstream>  
#include <iostream>  
#include <ctime>  
#include <cmath>  
#include <cstring>  
#include <cstdio>  
#include <time.h>  
#include <ctype.h>  
#include <string.h>  
#include <assert.h>  

using namespace std;

long long n, m, z, l;


int main()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%lld%lld%lld%lld", &n, &m, &z, &l);
        long long ans = 0;
        long long tmp = 0;
        for (int i = 2; i <= n;i++)
        {
            tmp = (tmp *m + z) % l;
            ans ^= (2 * tmp);
        }
        printf("%lld
",ans);
    }
    return 0;
}