MZL's xor

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1264    Accepted Submission(s): 784 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 题意：所有 ( Ai + Aj )( 1≤i,j≤n n)的异或值，最后只剩下2*a[i]的异或

code

#include <iostream> #include<cstdio> using namespace std; #define MAXN 500010 #define LL long long LL a[MAXN]; int main() { int t,m,n,z,l; 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=0; for(int i=1;i<=n;i++) { ans=ans^(2*a[i]); } printf("%lld\n",ans); } return 0; }