Codeforces 424C(异或)

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最新推荐文章于 2021-02-23 10:23:44 发布

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分类专栏：暴力模拟思维水题规律

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本文介绍了一种特殊的魔法公式计算方法，该方法应用于一系列正整数的处理中，通过使用位运算和特定的数学操作来计算最终的Q值。文章提供了一个C++实现示例，展示了如何读取输入并进行高效计算。

Magic Formulas

Time Limit: 2000MS

Memory Limit: 262144KB

64bit IO Format: %I64d & %I64u

Submit Status

Description

People in the Tomskaya region like magic formulas very much. You can see some of them below.

Imagine you are given a sequence of positive integer numbers p₁, p₂, ..., p_n. Lets write down some magic formulas:

$\text{[math]}$ $\text{[math]}$

Here, "mod" means the operation of taking the residue after dividing.

The expression $\text{[math]}$ means applying the bitwise xor (excluding "OR") operation to integers x and y. The given operation exists in all modern programming languages. For example, in languages C++ and Java it is represented by "^", in Pascal — by "xor".

People in the Tomskaya region like magic formulas very much, but they don't like to calculate them! Therefore you are given the sequence p, calculate the value of Q.

Input

The first line of the input contains the only integer n (1 ≤ n ≤ 10⁶). The next line contains n integers: p₁, p₂, ..., p_n (0 ≤ p_i ≤ 2·10⁹).

Output

The only line of output should contain a single integer — the value of Q.

Sample Input

Input

3
1 2 3

Output

Source

Codeforces Round #242 (Div. 2)

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
typedef long long LL;
int a[1000005],b[1000005];
int main()
{
    int i,j,t,len,n,s,ans;
    while(scanf("%d",&n)!=EOF)
    {
        scanf("%d",&t);
        ans=t;
        for(i=2;i<=n;i++)
        {
            scanf("%d",&t);
            ans^=t;
        }
        b[0]=0;
        for(i=1;i<=n;i++)
        {
            b[i]=b[i-1]^i;
            int tmp=n/i;
            if(tmp&1)
            ans^=b[i-1];
            if(n%i)
            ans=ans^b[n%i];
        }
        printf("%d\n",ans);
    }
    return 0;
}