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  • 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)(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

     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
    ", sum);
    45     }
    46 }
    View Code
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  • 原文地址:https://www.cnblogs.com/cdyboke/p/4875792.html
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