2015 Multi-University Training Contest 5 - 1002 MZL's xor-优快云博客

本博客探讨了MZL在XOR运算上的问题，包括数组生成规则、求解XOR运算结果的方法及实现代码。

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MZL's xor

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0

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

题意：

Ai=(Ai−1∗m+z)modl ，A1=0。求出所有任意两个Ai or Aj。1<=i,j<=n。

题解：

当i!=j时，必有两组Ai,Aj,Aj,Ai。他们的和相同，xor得0。0异或任何数为任何数本身。所以所有i!=j都不需要考虑只需求出当i==j时的结果。需要注意的是求出的Ai int可能会放不下。故开Long long 数组。

参考代码：

#include<stdio.h>
#define M 1000001
long long a[M];
int main()
{
	long long  t,n,m,z,l,sum;
	scanf("%lld",&t);
	while(t--)
	{
	    sum=0;
		scanf("%lld%lld%lld%lld",&n,&m,&z,&l);
        a[1]=0;
        for(int i=2;i<=n;i++)
        	a[i]=(a[i-1]*m+z)%l;
        for(int i=2;i<=n;i++)
            sum^=2*a[i];
        printf("%lld\n",sum);
	}
}