hdu 5683 zxa and xor 暴力

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⋅104,1≤m≤2⋅104,0≤ai,y≤109,1≤x≤n,1≤∑n,∑m≤105

 

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

Hint

After the first called function, this sequence is $\{4,2,3\}$, and $(4+2)\otimes(4+3)\otimes(2+3)=4$.

After the second called function, this sequence is $\{4,5,3\}$ and $(4+5)\otimes(4+3)\otimes(5+3)=6$.

After the third called function, this sequence is $\{4,5,6\}$ and $(4+5)\otimes(4+6)\otimes(5+6)=8$.

思路：通过位运算一些简单性质，暴力；（居然不会超时。。。）

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mod 1000000009
#define inf 999999999
#define esp 0.00000000001
//#pragma comment(linker, "/STACK:102400000,102400000")
int scan()
{
    int res =  , ch ;
    while( !( ( ch = getchar() ) >= '' && ch <= '' ) )
    {
        if( ch == EOF ) return  <<  ;
    }
    res = ch - '' ;
    while( ( ch = getchar() ) >= '' && ch <= '' )
        res = res *  + ( ch - '' ) ;
    return res ;
}
int a[];
int main()
{
    int x,y,z,i,t;
    scanf("%d",&x);
    while(x--)
    {
        scanf("%d%d",&y,&z);
        for(i=;i<=y;i++)
        scanf("%d",&a[i]);
        int sum=;
        for(i=;i<=y;i++)
        for(t=i+;t<=y;t++)
        sum^=(a[i]+a[t]);
        while(z--)
        {
            int pos,change;
            scanf("%d%d",&pos,&change);
            for(i=;i<=y;i++)
            {
                if(i!=pos)
                sum^=(a[i]+change)^(a[i]+a[pos]);
            }
            printf("%d\n",sum);
            a[pos]=change;
        }
    }
    return ;
}