2015 Multi-University Training Contest 5 1002

MZL's xor

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 276    Accepted Submission(s): 203

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

 

题意：给出n,m,z,l得到Ai的递推式，求所有Ai+Aj的异或值

分析：所有异或的话(Ai+Aj)和(Aj+Ai)会抵消，最后是对所有的2*Ai(1<= i <=n)进行异或

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<string>
#include<iostream>
#include<cstring>
#include<cmath>
#include<stack>
#include<queue>
#include<vector>
#include<map>
#include<stdlib.h>
#include<algorithm>
#define LL __int64
int main()
{
    int n,m,z,l,kase;
    scanf("%d",&kase);
    while(kase--)
    {
        scanf("%d %d %d %d",&n,&m,&z,&l);
        LL num=0,ans=0;
        for(int i=2;i<=n;i++)
        {
            num=(num*m+z)%l;
            ans=ans^2*num;
        }
        printf("%I64d
",ans);
    }
    return 0;
}