Codeforces 424 C Magic Formulas

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Time Limit:2000MS     Memory Limit:262144KB    64bit IO Format:%I64d & %I64u

SubmitStatus Practice CodeForces 424C

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 p_{1,p2, ...,pn. 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 of Q.

Input

The first line of the input contains the only integer n (1?≤?n?≤?10^{6). The next line containsn integers: p_{1,?p2,?...,?pn (0?≤?pi?≤?2·109).}}

Output

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

Sample Input

Input

31 2 3

Output

3

題意：如題。

思路：縱向分析，先是p1^……^pn。。。不說了，貼個網址，我覺得說的肯定比我清楚。http://www.tuicool.com/articles/InYrm2M

AC代碼：

#include <cstdio>#include <iostream>#include <algorithm>#include <cmath>#include <cstring>#include <stdlib.h>using namespace std;int p;int a[1000006];int main(){    int n;    scanf("%d",&n);    int ans=0;    for(int i=1;i<=n;i++){        scanf("%d",&p);        ans^=p;    }    for(int i=1;i<=n;i++){        a[i]=a[i-1]^i;        if(n%(2*i)!=0){            int x=n%(2*i);            if(x>=i){                ans^=a[i-1];                x-=i;            }            ans^=a[x];        }    }    printf("%d\n",ans);    return 0;}