[多校]MZL's xor

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,lA1=0,Ai=(Ai−1∗m+z) mod l1≤m,z,l≤5∗105,n=5∗105

 

Output

For every test.print the answer.

 

Sample Input

23 5 5 76 8 8 9

 

Sample Output

1416

     xor为“异或”，“异或”的规律为，1^1=0,0^0=0,1^0=1,0^1=1。这道题的题意是把A数组数据全部递推得到后，将Ai和Aj两两相加（注意！题中没说j>i或i>j。即存在A1+A2和A2+A1的情况）而Ai+Aj和Aj+Ai为相同值，相同值的异或为0,0对于其他值异或，值不会有改变。所以最后异或的数据只有当i==j时的数。（对于研究了一个小时为什么只需要对i==j情况异或的我，知道了真相以后震惊了）

#include <iostream>#include <cstdio>using namespace std;long long a[600010];int main(){    int n,m,z,l,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;        }        long long ans=a[1]+a[1];        for(int i=2;i<=n;i++)            ans^=(2*a[i]);        printf("%lld\n",ans);    }    return 0;}