F. Daniel and Spring Cleaning

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
While doing some spring cleaning, Daniel found an old calculator that he loves so much. However, it seems like it is broken. When he tries to compute 1+3 using the calculator, he gets 2 instead of 4. But when he tries computing 1+4, he gets the correct answer, 5. Puzzled by this mystery, he opened up his calculator and found the answer to the riddle: the full adders became half adders!

So, when he tries to compute the sum a+b using the calculator, he instead gets the xorsum a⊕b (read the definition by the link: https://en.wikipedia.org/wiki/Exclusive_or).

As he saw earlier, the calculator sometimes gives the correct answer. And so, he wonders, given integers l and r, how many pairs of integers (a,b) satisfy the following conditions:
a+b=a⊕b
l≤a≤r
l≤b≤r
However, Daniel the Barman is going to the bar and will return in two hours. He tells you to solve the problem before he returns, or else you will have to enjoy being blocked.

Input
The first line contains a single integer t (1≤t≤100) — the number of testcases.

Then, t lines follow, each containing two space-separated integers l and r (0≤l≤r≤109).

Output
Print t integers, the i-th integer should be the answer to the i-th testcase.

Example
inputCopy
3
1 4
323 323
1 1000000
outputCopy
8
0
3439863766
Note
a⊕b denotes the bitwise XOR of a and b.

For the first testcase, the pairs are: (1,2), (1,4), (2,1), (2,4), (3,4), (4,1), (4,2), and (4,3).

我们对条件：a+b=a⊕b 。可以想到，化成二进制之后，a，b不能有同一位都为1，不然肯定不相等，否则一定相等。

那么我们就可以数位dp了，令solve(x,y) 为a从0到x，b从0到y的方案数，然后就能二维容斥算出来了。

AC代码：

#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define int long long
using namespace std;
int T,l,r,dp[40][2][2],a[40],b[40];
int dfs(int pos,int l1,int l2){
	if(!pos)	return 1;
	if(dp[pos][l1][l2]!=-1)	return dp[pos][l1][l2];
	int up1=l1?a[pos]:1;
	int up2=l2?b[pos]:1,res=0;
	for(int i=0;i<=up1;i++){
		for(int j=0;j<=up2;j++){
			if((i&j))	continue;
			res+=dfs(pos-1,l1&&i==up1,l2&&j==up2);
		}
	}
	dp[pos][l1][l2]=res;
	return res;
}
int solve(int l,int r){
	if(l==-1||r==-1)	return 0;
	int pos=0; memset(dp,-1,sizeof dp);
	for(;l||r;l>>=1,r>>=1)	a[++pos]=l&1,b[pos]=r&1;
	return dfs(pos,1,1);
}
signed main(){
	cin>>T; memset(dp,-1,sizeof dp);
	while(T--){
		cin>>l>>r;
		cout<<solve(r,r)-solve(l-1,r)*2+solve(l-1,l-1)<<endl;
	}
	return 0;
}