Codeforces 424C(异或)
2016-05-24 10:16
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Magic Formulas
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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, ..., 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.
Input
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).
Output
The only line of output should contain a single integer — the value of Q.
Sample Input
Input
Output
Source
Codeforces Round #242 (Div. 2)
Time Limit: 2000MS | Memory Limit: 262144KB | 64bit IO Format: %I64d & %I64u |
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, ..., 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.
Input
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).
Output
The only line of output should contain a single integer — the value of Q.
Sample Input
Input
3 1 2 3
Output
3
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; }
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