MZL's xor

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 800    Accepted Submission(s): 518

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

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

2

3 5 5 7

6 8 8 9

14

16

(Ai+Aj)^(Aj+Ai)=0 (i≠j)

然后注意开long long 否则 ai*m时会爆

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<algorithm>

#include<functional>

#include<iostream>

#include<cmath>

#include<cctype>

#include<ctime>

using namespace std;

#define For(i,n) for(int i=1;i<=n;i++)

#define Fork(i,k,n) for(int i=k;i<=n;i++)

#define Rep(i,n) for(int i=0;i<n;i++)

#define ForD(i,n) for(int i=n;i;i--)

#define RepD(i,n) for(int i=n;i>=0;i--)

#define Forp(x) for(int p=pre[x];p;p=next[p])

#define Forpiter(x) for(int &p=iter[x];p;p=next[p])

#define Lson (x<<1)

#define Rson ((x<<1)+1)

#define MEM(a) memset(a,0,sizeof(a));

#define MEMI(a) memset(a,127,sizeof(a));

#define MEMi(a) memset(a,128,sizeof(a));

#define INF (2139062143)

#define F (100000007)

#define MAXN (5000000+10)

typedef long long ll;

ll mul(ll a,ll b){return (a*b)%F;}

ll add(ll a,ll b){return (a+b)%F;}

ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}

void upd(ll &a,ll b){a=(a%F+b%F)%F;}

ll a[MAXN];

ll n,m,z,l;

int main()

{

//	freopen("B.in","r",stdin);

	int T;cin>>T;

	while(T--)

	{

		cin>>n>>m>>z>>l;

		a[1]=0;

		Fork(i,2,n) a[i]=(a[i-1]*m+z)%l;

		ll s=0;

		For(i,n) s=s^(2*a[i]);

		cout<<s<<endl;

	} 

	return 0;

}