HDU 5683 zxa and xor （模拟）

最新推荐文章于 2016-09-01 21:21:47 发布

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最新推荐文章于 2016-09-01 21:21:47 发布

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分类专栏：模拟

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本文介绍了一种使用模拟算法解决序列XOR运算的问题，通过更新元素值并重新计算所有可能的两两元素之和的XOR结果来求解每次变更后的序列状态。

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思路：这题居然能模拟...

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e4+6;
int a[maxn],ans,n,m;
void solve()
{
    scanf("%d%d",&n,&m);
    ans=0;
    for(int i=1;i<=n;i++)scanf("%d",&a[i]);
    for(int i=1;i<=n;i++)for(int j=i+1;j<=n;j++)
        ans^=(a[i]+a[j]);
    while(m--)
    {
        int x,y;scanf("%d%d",&x,&y);
        for(int i=1;i<x;i++)ans^=(a[x]+a[i])^(y+a[i]);
        for(int i=x+1;i<=n;i++)ans^=(a[x]+a[i])^(y+a[i]);
        a[x]=y;
        printf("%d\n",ans);
    }
}
int main()
{
    int t;scanf("%d",&t);
    while(t--)solve();
    return 0;
}

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

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 $(4+5)\otimes(4+6)\otimes(5+6)=8$.

Source

BestCoder Round #83