codeforces 424C. Magic Formulas （异或规律）

链接：http://codeforces.com/problemset/problem/424/C

题目：

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:

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

The expression means applying the bitwisexor (excluding "OR") operation to integersx 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 sequencep, calculate the value ofQ.

Input

The first line of the input contains the only integer n (1 ≤ n ≤ 10⁶). The next line containsn 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 test(s)

Input

31 2 3

Output

3

分析：

i mod(1)的归为一类，i mod(2)的归为一类……

这样一列一列地研究，发现这是周期性的。利用数字本身和本身异或得0的事实，得出这样的结论：

设

当i/j是偶数时
当i/j是奇数时

如果觉得我说的不清楚，你带入i=14, j=3和j=16, j=3来试试，你会懂得。

#include <iostream>#include <cstdio>using namespace std;const int N=1e6+10;int dp[N];int main(){    for(int i=1;i<N;i++){        dp[i]=dp[i-1]^i;    }    int n,p;    while(cin>>n){        int ans=0;        for(int i=0;i<n;i++){            scanf("%d",&p);            ans=ans^p;        }        for(int i=2;i<=n;i++){            int r=n%i;            int div=n/i;            if(div&1) ans=ans^dp[i-1];            ans=ans^dp[r];        }        printf("%d\n",ans);    }    return 0;}