HDU 5344 MZL's xor （多校）[补7月28]

MZL's xor

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 249    Accepted Submission(s): 187

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

 

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题意理解了半天，他说要把所有的（Ai+Aj）拿来异或，Ai+Aj和Aj+Ai都算的，所以只要i和j
不一样，(Ai+Aj)XOR(Aj+Ai)一定是0，这些都不用考虑了，只有i=j的才考虑，所以把所有的Ai拿来乘以2，再异或就可以了。

#include<queue>
#include<math.h>
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define N 1234567
#define M 1234

long long a[N];
long long  b[N];
int ma,n,num;
int main()
{
    a[1]=0;
    int t;cin>>t;
    while(t--)
    {
        int n,m,z,l;
        scanf("%d%d%d%d",&n,&m,&z,&l);
        for(int i=2;i<=n;i++)
            a[i]=(a[i-1]*m+z)%l;

        int c=a[1]*2;
        for(int i=1;i<=n;i++)
        {
            c=c^a[i]*2;
        }
        printf("%d
",c);
    }
    return 0;
}