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

简单题，异或运算会互相抵消，所以只剩下ai+ai了

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=500005;
typedef long long ll;
ll a[maxn],T,n,m,z,mod,ans,sum;

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