hdu 5344 MZL's xor（数学之异或）

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

Source

2015 Multi-University Training Contest 5

找到了规律就好办了，发现对称性，相同的数异或后等于0，所以最后剩下自身*2来异或，注意要用long long

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<queue>
 6 #include<cmath>
 7 #include<stdlib.h>
 8 #include<map>
 9 using namespace std;
10 #define ll long long
11 #define N 600000
12 ll n,m,z,l;
13 ll a[N];
14 int main()
15 {
16     int t;
17     scanf("%d",&t);
18     while(t--){
19         scanf("%I64d%I64d%I64d%I64d",&n,&m,&z,&l);
20         a[1]=0;
21         for(int i=2;i<=n;i++){
22             a[i]=(a[i-1]*m+z)%l;
23         }
24         ll ans=0;
25         for(int i=1;i<=n;i++){
26             ans=ans^(a[i]+a[i]);
27         }
28         printf("%I64d\n",ans);
29     }
30     return 0;
31 }

hdu 5344 MZL's xor（数学之异或）