HDU 5344(MZL's xor-(ai+aj)的异或和)

MZL's xor

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 800 Accepted Submission(s): 518



[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]Author[/align]
SXYZ

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

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(Ai+Aj)^(Aj+Ai)=0 (i≠j)

然后注意开long long 否则 ai*m时会爆

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (5000000+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
ll a[MAXN];
ll n,m,z,l;
int main()
{
//	freopen("B.in","r",stdin);
int T;cin>>T;
while(T--)
{
cin>>n>>m>>z>>l;
a[1]=0;
Fork(i,2,n) a[i]=(a[i-1]*m+z)%l;
ll s=0;
For(i,n) s=s^(2*a[i]);
cout<<s<<endl;
}

return 0;
}