【HDU 5344 MZL's xor】+ 异或_hdu5344-优快云博客

本文链接：https://blog.youkuaiyun.com/WYK1823376647/article/details/60882657

本文介绍了一道关于数组XOR运算的编程题，重点在于理解如何通过算法快速计算两个不同数组元素组合相加后的XOR值。文章提供了一段AC代码作为解决方案。

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MZL’s xor

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1259 Accepted Submission(s): 780

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

B 为所有a[i] + a[j] 的值组成的集合，所以有a[i] + a[j] 必有a[j] + a[i],且相同值异或为0，所以只用考虑 2 * a[i]的异或

AC代码：

#include<cstdio>
typedef long long LL;
int main()
{
    int T,n,m,z,l;
    scanf("%d",&T);
    while(T--){
        scanf("%d %d %d %d",&n,&m,&z,&l);
        LL ans = 0,a = 0; while(n--) ans ^= a * 2,a = (a * m + z) % l;
        printf("%lld\n",ans);
    }
    return 0;
}