HDOJ-----5344---MZL's xor---思维

MZL's xor

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1279    Accepted Submission(s): 795

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...xorBn

 

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)modl
1≤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

给出A1，以及Ai的通项Ai=(Ai−1∗m+z) mod  l，可以求出数列A，求所有的（Ai+Aj）^（Ap+Aq）                    0 < i，j，p，q <= n

      A1     A2

A1   1       2

A2   2       3

首先考虑上面这个表的性质，对角线两侧完全对称即会出现（A1+A2）^（A2+A1）

考虑异或的性质，二进制位相同为0，不同为1，a^a = 0

所以求所有的（Ai+Aj）^（Ap+Aq），对角线两侧对称，全部抵消为0，最终是求对角线（Ai+Ai）^（A(i+1)+A(i+1)）^ …… ^（An+An)

最终即2*（Ai ^ …… ^ An）

#include<cstdio>#include<iostream>#include<cstring>#define LL long longusing namespace std;int main(){int t, n, m, z, l;cin >> t;while(t--){cin >> n >> m >> z >> l;LL a, ans;ans = a = 0;for(int i = 0; i < n; i++){ans ^= a;a = (a*m+z)%l;}cout << 2*ans << endl;}return 0;}