HDU 5344(MZL's xor-(ai+aj)的异或和)
原创
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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
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
Sample Output
Author
SXYZ
Source
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;
}