HDU 5344 MZL's xor （多校）[补7月28]

MZL's xor

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 249 Accepted Submission(s): 187


[align=left]Problem Description[/align]
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

[align=left]Input[/align]
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

[align=left]Output[/align]
For every test.print the answer.

[align=left]Sample Input[/align]

2
3 5 5 7
6 8 8 9

[align=left]Sample Output[/align]

14
16

[align=left]Source[/align]
2015 Multi-University Training Contest 5

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题意理解了半天，他说要把所有的（Ai+Aj）拿来异或，Ai+Aj和Aj+Ai都算的，所以只要i和j
不一样，(Ai+Aj)XOR(Aj+Ai)一定是0，这些都不用考虑了，只有i=j的才考虑，所以把所有的Ai拿来乘以2，再异或就可以了。

#include<queue>
#include<math.h>
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define N 1234567
#define M 1234

long long a
;
long long  b
;
int ma,n,num;
int main()
{
a[1]=0;
int t;cin>>t;
while(t--)
{
int n,m,z,l;
scanf("%d%d%d%d",&n,&m,&z,&l);
for(int i=2;i<=n;i++)
a[i]=(a[i-1]*m+z)%l;

int c=a[1]*2;
for(int i=1;i<=n;i++)
{
c=c^a[i]*2;
}
printf("%d\n",c);
}
return 0;
}