#题目：GCD XOR UVA - 12716

题目描述

Given an integer N, find how many pairs (A, B) are there such that: gcd(A, B) = A xor B where
1 ≤ B ≤ A ≤ N.
Here gcd(A, B) means the greatest common divisor of the numbers A and B. And A xor B is the
value of the bitwise xor operation on the binary representation of A and B.

input
The first line of the input contains an integer T (T ≤ 10000) denoting the >number of test cases. The
following T lines contain an integer N (1 ≤ N ≤ 30000000).
output
For each test case, print the case number first in the format, ‘Case X:’ (here, X is the serial of the
input) followed by a space and then the answer for that case. There is no new-line between cases.
Sample Input
2
7
20000000
Sample Output
Case 1: 4
Case 2: 34866117

题意：

多组数据输入，每组数据输入一个n，问在小于n的数字里面能找到多少对数满足gcd（a,b)==a^b(大于b)

解题思路：

根据异或的性质我们可以发现，a^gcd(a,b)=b;
而gcd(a,b)又是a的约数，所以我们可以枚举c，就可以像筛素数那样做，然后在暴力小数据打表的时候发现gcd(a,b)=a-b，证明详见紫书318页。所以我们在筛的时候就找a^(a-b)是否等于b就行

代码

#include <iostream>
using namespace std;

int gcd (int a,int b)
{
    if(b==0)return a;
    else return gcd(b,a%b);
}
int a[300000005];

void init()
{
    a[1]=0;
    a[2]=0;

    for(int i=1;i<300000007;i++)
    {
        for(int j=i+i;j<30000007;j+=i)
        {
            int tmp = j-i;
            if((j^tmp) == i)
            {
                a[j]++;
            }

        }
        a[i]+=a[i-1];
    }


}

int main()
{
   init();
   int t;
   cin>>t;
   int flg=1;
   while(t--)
   {
    int n;
    cin>>n;
    int ans=a[n];
    cout<<"Case "<<flg++<<": "<<ans<<endl;

   }
   return 0;
}

总结：

要多去想规律，比如说看到异或就要想到：如果a^b=c那么a^c=b,然后还要结合打表。

转载于:https://www.cnblogs.com/IAMjm/p/9429146.html