MZL‘s xor

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1509    Accepted Submission(s): 619

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

水题

两个相等的数异或为0

先用它给的公式算出 a[1..n]来

然后 答案就是（2倍的a[i] ）的异或。。。

/*************************************************************************
    > File Name: code/multi/#5/1002.cpp
    > Author: 111qqz
    > Email: [email protected]
    > Created Time: 2015年08月04日 星期二 13时51分14秒
 ************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define y0 abc111qqz
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define tm crazy111qqz
#define lr dying111qqz
using namespace std;
#define REP(i, n) for (int i=0;i<int(n);++i)
typedef long long LL;
typedef unsigned long long ULL;
const int inf = 0x7fffffff;
const int N=5E5+7;
LL a[N];
LL ans;
int n,m,z,l;
int main()
{
    int T;
    cin>>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;
    }

    ans = 0;
    for ( int i = 1 ;i <= n;  i++ )
    {
        ans = ans ^ (a[i]+a[i]);
    }
    printf("%lld\n",ans);
    }

    return 0;
}

2015多校 #5 1002 MZL's xor