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

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

题意：求所有a[i]+a[j] 异或值，因为a[i]+a[j] 和a[j]+a[i] 会异或掉 所以只需考虑a[i]+a[i]

代码：

#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\n",ans);
    }
    return 0;
}