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hdu 5344 MZL's xor

編輯：關於C++

MZL's xor

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 310 Accepted Submission(s): 225

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

題意：求所有a[i]+a[j] 異或值，因為a[i]+a[j] 和a[j]+a[i] 會異或掉所以只需考慮a[i]+a[i]

代碼：

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using namespace std;

long long n, m, z, l;


int main()
{
    int t;
    scanf(%d, &t);
    while (t--)
    {
        scanf(%lld%lld%lld%lld, &n, &m, &z, &l);
        long long ans = 0;
        long long tmp = 0;
        for (int i = 2; i <= n;i++)
        {
            tmp = (tmp *m + z) % l;
            ans ^= (2 * tmp);
        }
        printf(%lld
,ans);
    }
    return 0;
}