MZL's xor

MZL's xor

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/65536K (Java/Other)

Total Submission(s) : 1   Accepted Submission(s) : 1

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 ($A_i$+$A_j$)($1\leq i,j \leq n$)
The xor of an array B is defined as $B_1$ xor $B_2$...xor $B_n$

 

 

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:$n$,$m$,$z$,$l$ $A_1=0$,$A_i=(A_{i-1}*m+z)$ $mod$ $l$ $1\leq m,z,l \leq 5*10^5$,$n=5*10^5$

 

 

Output

For every test.print the answer.

 

 

Sample Input

2 3 5 5 7 6 8 8 9

 

 

Sample Output

14 16

 题解：就是A数组可以得出，B数组是A[i]+A[j]组成的数组让求B1^B2^.......^Bn的值；

代码：

#include<stdio.h>
#include<string.h>
const int MAXN=5*1e5+10;
long long A[MAXN],B[MAXN];
int main(){
    int T,n,m,z,l;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d%d%d",&n,&m,&z,&l);
        A[0]=0;
        long long ans=0;
        for(int i=1;i<n;i++){
            A[i]=(A[i-1]%l*m%l+z%l)%l;
            B[i]=A[i]+A[i];//因为相同的元素异或结果为0；所以A[i]+A[j] | A[j]+A[i]等于0；只剩相同元素之和的异或； 
            ans^=B[i];
        }
        printf("%lld\n",ans);
    }
    return 0;
}