MZL’s xor
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1259 Accepted Submission(s): 780
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 2015 Multi-University Training Contest 5
B 为 所有a[i] + a[j] 的值组成的集合，所以有a[i] + a[j] 必有a[j] + a[i],且相同值异或为0，所以只用考虑 2 * a[i]的异或
AC代码：
#include<cstdio> typedef long long LL; int main() { int T,n,m,z,l; scanf("%d",&T); while(T--){ scanf("%d %d %d %d",&n,&m,&z,&l); LL ans = 0,a = 0; while(n--) ans ^= a * 2,a = (a * m + z) % l; printf("%lld\n",ans); } return 0; }