2298. 【noip普及组2T4】异或 
(File IO): input:gcdxor.in output:gcdxor.out

题目描述

SarvaTathagata是个神仙,一天他在研究数论时,书上有这么一个问题:求不超过n两两的数的gcd。
SarvaTathagata这么神仙的人当然觉得这个是sb题啦。学习之余,他还发现gcd的某一个特别好的性质:如果有两个数i,j满足gcd(i,j)=i^j(这里的^为c++中的异或)的话,那么这两个数组成的数对(i,j)就是一个nb的数对(这里认为(i,j)和(j,i)为相同的,并不需要计算2次)。
当然,SarvaTathagata并不会只满足于判断一个数对是否nb,他还想知道满足两个数都是不超过n并且nb的数对有多少个。
由于SarvaTathagata实在是太神仙了,他认为这种题实在是太简单了。于是他找到了你,看看你是否能解决这个问题。
 

输入

共一行一个整数n,含义如题所述。

输出

一行一个整数,表示nb的数对的个数。
 

样例输入

样例输入1
12
样例输入2
123456

样例输出

样例输出1
8
样例输出2
214394
 

数据范围限制

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" alt=" " />
 

提示

样例1中共有八对,分别是: {1,3},{1,5},{1,7},{1,9},{2,6},{1,11},{2,10},{4,12}。
 

伪思路:

直接判断,二话不说。

假的CODE:

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int n,ans=;
int gcd(int x,int y)
{
if(y==)
return x;
else
gcd(y,x%y);
}
int main()
{
cin>>n;
for(int i=;i<=n;i++)
{
for(int j=i+;j<=n;j++)
{
if(gcd(i,j)==(i^j))
ans++;
}
}
cout<<ans;
}

很可惜,第二个样例就已经卡的动不了,哈哈哈

(ノ=Д=)ノ┻━┻

真~思路(不严谨的思考):

我们用原方案是不可以的,毫无疑问是这个数据太坑爹了!!!最郁闷的还是这个数据解释是假的,假的,假的!!!

经过一番痛苦的思考(快乐的偷窥)我们可以得出gcd(a,b)是一定小于等于abs(a-b)的。证明:易得证【滑稽】毫无疑问,我们知道a+b大于等于1;不妨设系数k,则有ak+bk大于等于k;将其转换可以得出gcd(a,b)<=abs(a-b)接下了我们确定abs(a-b)的上限,结合题意很容易猜出是i亦或j。证明:将两个东西转换成二进制。两者大多相同,但当0在上1在下时,相减会退位而亦或会想加。

真~CODE:

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int n,ans=;
int main()
{
cin>>n;
for(int i=;i<=n;i++)
{
for(int j=*i;j<=n;j+=i)
{
if(j-i==(i^j))
ans++;
}
}
cout<<ans;
}

完结撒花!!!

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