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这个题就是求出给的公式的结果。
只要知道异或运算满足交换律跟结合律就行了，之后就是化简公式。
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<climits>
#include<list>
#include<iomanip>
#include<stack>
#include<set>
using namespace std;
int psum[1000010];
int main()
{
int n;
cin>>n;
int ans=0;
for(int i=0;i<n;i++)
{
int t;
scanf("%d",&t);
ans^=t;
}
for(int i=1;i<n;i++)
psum[i]=i^psum[i-1];
for(int i=1;i<=n;i++)
{
int t=n/i;
if(t&1)
ans^=psum[i-1];
t=n%i;
ans^=psum[t];
}
cout<<ans;
}
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, ..., pn. Lets write down some magic formulas:
Here, "mod" means the operation of taking the residue after dividing.
The expression 
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.
The first line of the input contains the only integer n (1?≤?n?≤?106). The next line contains n integers: p1,?p2,?...,?pn (0?≤?pi?≤?2·109).
The only line of output should contain a single integer — the value of Q.
3 1 2 3
3
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原文地址：http://blog.csdn.net/stl112514/article/details/47146013
