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


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)(1i,jn)
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=(Ai1m+z) mod l
1m,z,l5105,n=5105
 

 

Output
For every test.print the answer.
 

 

Sample Input
2 3 5 5 7 6 8 8 9
 

 

Sample Output
14 16
 

 

Source
 

 

Recommend
wange2014   |   We have carefully selected several similar problems for you:  5352 5351 5350 5349 5348
 
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。
 1 #include<queue>
 2 #include<math.h>
 3 #include<stdio.h>
 4 #include<string.h>
 5 #include<iostream>
 6 #include<algorithm>
 7 using namespace std;
 8 #define N 1234567
 9 #define M 1234
10 
11 long long a[N];
12 long long  b[N];
13 int ma,n,num;
14 int main()
15 {
16     a[1]=0;
17     int t;cin>>t;
18     while(t--)
19     {
20         int n,m,z,l;
21         scanf("%d%d%d%d",&n,&m,&z,&l);
22         for(int i=2;i<=n;i++)
23             a[i]=(a[i-1]*m+z)%l;
24 
25         int c=a[1]*2;
26         for(int i=1;i<=n;i++)
27         {
28             c=c^a[i]*2;
29         }
30         printf("%d\n",c);
31     }
32     return 0;
33 }

 

posted @ 2015-08-04 21:03  wmxl  阅读(269)  评论(0编辑  收藏  举报