hdoj 5433 MZL's xor ( 异或性质)

MZL's xor

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1215    Accepted Submission(s): 745

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

 

异或性质：a^a=0   ,  a^0=a  .

题解：发现求 (A1+A1)^(A1+A2)^.......(A2+A1)^(A2+A2).......(An+An),最后(Ai+Aj)^(Aj+Ai)=0,    sum=0^(2*A1)......^(2*An);

异或 ^ 是运算符。

代码：

#include<cstdio>int a[1000000];int main(){int t,n,m,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]=((long long)a[i-1]*m+z)%l;//用long long }int sum=0;for(int i=1;i<=n;i++){sum=sum^(2*a[i]);}printf("%d\n",sum);}return 0;}