Problem Description
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
简单题,异或运算会互相抵消,所以只剩下ai+ai了
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=500005;
typedef long long ll;
ll a[maxn],T,n,m,z,mod,ans,sum;
int main()
{
scanf("%lld",&T);
while (T--)
{
scanf("%lld%lld%lld%lld",&n,&m,&z,&mod);
ans=a[1]=0;
for (int i=2;i<=n;i++)
{
a[i]=(a[i-1]*m+z)%mod;
ans=ans^(a[i]+a[i]);
}
printf("%d\n",ans);
}
return 0;
}