HDU5344——数学题——MZL's xor

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

 

Source

2015 Multi-University Training Contest 5

 

/*
根据式子可以推出公式为所有值×2的异或和
*/
/************************************************
Author        :powatr
Created Time  :2015-8-5 22:08:30
File Name     :b.cpp
************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int MAX = 1e6 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;

ll a[MAX];
int main(){
    int n, m, z, l;
    int T;
    scanf("%d", &T);
    while(T--){
    scanf("%d%d%d%d", &n, &m, &z, &l);
    a[1] = 0;
        for(int i = 2; i <= n; i++)
            a[i] = (a[i-1]*m + z)%l;
    ll sum = 0;
    for(int i = 1; i <= n; i++)
        sum ^= 1ll*(a[i] << 1);
  printf("%I64d\n", sum);
    }
  return 0;
}