hdu-5683 zxa and xor (位运算)

题目链接：

zxa and xor

Time Limit: 16000/8000 MS (Java/Others)   

 Memory Limit: 65536/65536 K (Java/Others)

Problem Description

zxa had a great interest in exclusive disjunction(i.e. XOR) recently, therefore he took out a non-negative integer sequence a1,a2,⋯,an of length n.

zxa thought only doing this was too boring, hence a function funct(x,y) defined by him, in which ax would be changed into y irrevocably and then compute ⊗1≤i<j≤n(ai+aj) as return value.

zxa is interested to know, assuming that he called such function m times for this sequence, then what is the return value for each calling, can you help him?

tips:⊗1≤i<j≤n(ai+aj) means that (a1+a2)⊗(a1+a3)⊗⋯⊗(a1+an)⊗(a2+a3)⊗(a2+a4)⊗⋯⊗(a2+an)⊗⋯⊗(an−1+an).

Input

The first line contains an positive integer T, represents there are T test cases.

For each test case:

The first line contains two positive integers n and m.

The second line contains n non-negative integers, represent a1,a2,⋯,an.

The next m lines, the i-th line contains two non-negative integers x and y, represent the i-th called function is funct(x,y).

There is a blank between each integer with no other extra space in one line.

1≤T≤1000,2≤n≤2⋅10^4,1≤m≤2⋅10^4,0≤ai,y≤10^9,1≤x≤n,1≤∑n,∑m≤10^5

Output

For each test case, output in m lines, the i-th line a positive integer, repersents the return value for the i-th called function.

Sample Input

1

3 3

1 2 3

1 4

2 5

3 6

Sample Output

4

6

8

题意：

给一个数列,每次把a[x]变成y,问每次的fun是多少;

思路：

先算一次fun,再在每次变换的时候运用a^a=0的性质把a[x]+a[i]再异或一边,然后把y+a[i]异或和上去就好了;

AC代码：

//#include <bits/stdc++.h>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
#define Riep(n) for(int i=1;i<=n;i++)
#define Riop(n) for(int i=0;i<n;i++)
#define Rjep(n) for(int j=1;j<=n;j++)
#define Rjop(n) for(int j=0;j<n;j++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
const LL mod=1e9+;
const double PI=acos(-1.0);
const int inf=0x3f3f3f3f;
const int N=2e4+;
int n,m;
int a[N],ans;
int get_ans()
{
     ans=;
    for(int i=;i<=n;i++)
    {
        for(int j=i+;j<=n;j++)
        {
            int sum=a[i]+a[j];
            ans^=sum;
        }
    }
}
int fun(int fx,int fy)
{
    for(int i=;i<=n;i++)
    {
       if(i!=fx) ans^=(a[fx]+a[i])^(fy+a[i]);
    }
    a[fx]=fy;
}
int main()
{
   int t;
   scanf("%d",&t);
   while(t--)
   {
       scanf("%d%d",&n,&m);
       for(int i=;i<=n;i++)
       {
           scanf("%d",&a[i]);
       }
       int x,y;
       get_ans();

       while(m--)
       {
           scanf("%d%d",&x,&y);
           fun(x,y);
           printf("%d\n",ans);
       }
   }
    return ;
}