hdoj 5344 MZL's xor (简单异或)

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

23 5 5 76 8 8 9

 

Sample Output

1416

题意：所有(Ai+Aj)(1≤i,j≤nn)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 longLL 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;}