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。