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
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 asB1
xor B2 ...xor
Bn
The xor of an array B is defined as
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
Each test case contains four integers:
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
Recommend
(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; }