hdu 5344 MZL's xor

　　

MZL's xor

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 911    Accepted Submission(s): 589

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

 

Author

SXYZ

 

Source

2015 Multi-University Training Contest 5

AxorA=0 Axor0=A

i!=j时（Ai+Aj）xor（Ai+Aj)=0,剩下(A1+A1)xor(A2+A2)xor...xor(An+An),而Ai=(Ai−1∗m+z) mod l是产生随机数的函数，在一定长度下会出现循环,循环部分又可以约掉

 #include<cstdio>
 #include<cstring>
 bool b[];
 int  a[];
 int main()
 {
     long long z, m, l;
     int n, T;
     scanf("%d", &T);
     while (T--)
     {
         scanf("%d%lld%lld%lld", &n, &m, &z, &l);
         int j, f, L, i;
         memset(b, , sizeof(b));
         L = ;
         b[] = true;
         a[] = ;
         for (j = ;; j++)
         {
             L = (m*L + z) % l;//随机数产生函数，会有循环出现
             if (b[L] == true) break;
             b[L] = true;
             a[j] = L;
         }
         int sum = ;
         for (f = ; f < j; f++)
         {
             if (a[f] == L) break;
             sum = sum ^ (a[f] << );
         }
         int cycle = j - f;
         int div = (n - f) / cycle;
         int mod = (n - f) % cycle;
         if (div & )
         {
             for (i = f + mod; i < j; i++)
                 sum = sum ^ (a[i] << );
         }
         else
         {
             for (i = f; i < f + mod; i++)
                 sum = sum ^ (a[i] << );
         }
         printf("%d\n", sum);
     }
 }