CodeForce 424C Magic Formulas
这个题就是求出给的公式的结果。
仅仅要知道异或运算满足交换律跟结合律即可了。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;
}
2 seconds
256 megabytes
standard input
standard output
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
CodeForce 424C Magic Formulas的更多相关文章
- Codeforce 424C Magic Formulas 找规律
题目链接:http://codeforces.com/contest/424/problem/C 题意:求Q值 思路:找规律 显然能够得到一个矩阵 把这个矩阵画出来就能发现一个横向的规律和一个主对角线 ...
- Codeforces 424 C. Magic Formulas
xor是满足交换律的,展开后发现仅仅要能高速求出 [1mod1....1modn],....,[nmod1...nmodn]的矩阵的xor即可了....然后找个规律 C. Magic Formulas ...
- codeforce-424C. Magic Formulas(数学)
C. Magic Formulas time limit per test:2 seconds memory limit per test:256 megabytes input stan ...
- Codeforces Round #242 (Div. 2) C. Magic Formulas
解题思路是: Q=q1^q2.......^qn = p1^p2......^pn^((1%1)^....(1%n))^((2%1)^......(2%n))^.... 故Q的求解过程分成两部分 第一 ...
- Codeforces Round #242 (Div. 2) C. Magic Formulas (位异或性质 找规律)
题目 比赛的时候找出规律了,但是找的有点慢了,写代码的时候出了问题,也没交对,还掉分了.... 还是先总结一下位移或的性质吧: 1. 交换律 a ^ b = b ^ a 2. 结合律 (a^b) ^ ...
- codeforce 606A - Magic Spheres
题意:a,b,c三种球,能把俩个一样的球变成另一颜色不一样的球.给你目标x,y,z,问能否经过变化至少达打目标. #include<iostream> #include<stdio. ...
- cf C. Magic Formulas
http://codeforces.com/contest/424/problem/C #include <cstdio> #include <cstring> #includ ...
- codeforces C. Magic Formulas 解题报告
题目链接:http://codeforces.com/problemset/problem/424/C 题目意思:给出 n 个数:p1, p2, ..., pn,定义: q1 = p1 ^ (1 mo ...
- Codeforces Round #242 (Div. 2) <A-D>
CF424 A. Squats 题目意思: 有n(n为偶数)个x和X,求最少的变换次数,使得X的个数为n/2,输出变换后的序列. 解题思路: 统计X的个数ans,和n/2比較,少了的话,须要把n/2- ...
随机推荐
- neo4j nosql图数据库学习
neo4j 文档:https://neo4j.com/docs/getting-started/current/cypher-intro/ 1.索引 # 给某类标签节点的属性添加索引 CREATE I ...
- python在leecode刷题-第一题和第七题
class Solution(object): def twoSum(self, nums, target): """ :type nums: List[int] num ...
- HDU 4906 Our happy ending
题意: Given a sequence a_1,a_2,...,a_n, if we can take some of them(each a_i can only be used once), a ...
- error C2440: “static_cast”: 无法从“LRESULT (__thiscall CTextProgressCtrl::* )(UINT,LPCTSTR)”转换为“LRESULT (__thiscall CWnd::* )(WPARAM,LPARAM)
转自原文 error C2440 “static_cast” 无法从“void (__thiscall C* )(void)... error C2440: “static_cast”: 无法从“LR ...
- Mac上配置 Ruby on Rails和Git
Ruby on Rails on Mac =============================================================================== ...
- MKVToolNix v8.2
32位版:http://pan.baidu.com/s/1i3s4gGd 64位版: http://pan.baidu.com/s/1gdvqbpp
- HBase源代码分析之HRegion上MemStore的flsuh流程(二)
继上篇<HBase源代码分析之HRegion上MemStore的flsuh流程(一)>之后.我们继续分析下HRegion上MemStore flush的核心方法internalFlushc ...
- FZU--2188--过河(bfs暴力条件判断)
过河I Time Limit: 3000MS Memory Limit: 32768KB 64bit IO Format: %I64d & %I64u Submit Status De ...
- node.js操作Cookie
node.js操作Cookie http://www.tuicool.com/articles/F3UF7n
- js垃圾回收机制理解
原理 找到不再被使用的变量,然后释放其占用的内存,但这个过程不是时时的,因为其开销比较大, 所以垃圾回收器会按照固定时间间隔周期性的执行 回收方式 a.标记清除 当变量进入环境时,将这个变量标记为“进 ...