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hdu 5344 MZL's xor

时间:2015-10-13 22:32:39      阅读:337      评论:0      收藏:0      [点我收藏+]

  

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)(1i,jn)
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=(Ai1m+z) mod l
1m,z,l5105,n=5105
 

 

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
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=(Ai1m+zmol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

技术分享
 1 #include<cstdio>
 2 #include<cstring>
 3 bool b[500005];
 4 int  a[500005];
 5 int main()
 6 {
 7     long long z, m, l;
 8     int n, T;
 9     scanf("%d", &T);
10     while (T--)
11     {
12         scanf("%d%lld%lld%lld", &n, &m, &z, &l);
13         int j, f, L, i;
14         memset(b, 0, sizeof(b));
15         L = 0;
16         b[0] = true;
17         a[0] = 0;
18         for (j = 1;; j++)
19         {
20             L = (m*L + z) % l;//随机数产生函数,会有循环出现
21             if (b[L] == true) break;
22             b[L] = true;
23             a[j] = L;
24         }
25         int sum = 0;
26         for (f = 0; f < j; f++)
27         {
28             if (a[f] == L) break;
29             sum = sum ^ (a[f] << 1);
30         }
31         int cycle = j - f;
32         int div = (n - f) / cycle;
33         int mod = (n - f) % cycle;
34         if (div & 1)
35         {
36             for (i = f + mod; i < j; i++)
37                 sum = sum ^ (a[i] << 1);
38         }
39         else
40         {
41             for (i = f; i < f + mod; i++)
42                 sum = sum ^ (a[i] << 1);
43         }
44         printf("%d\n", sum);
45     }
46 }
View Code

 

hdu 5344 MZL's xor

原文:http://www.cnblogs.com/cdyboke/p/4875792.html

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