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

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

 

Author

SXYZ

 

Source

 

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(Ai+Aj)^(Aj+Ai)=0 (i≠j)

然后注意开long long 否则 ai*m时会爆

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               =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<
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