F. Daniel and Spring Cleaning
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
input
3
1 4
323 323
1 1000000
output
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).
题意
给你l，r；问你[l,r]中有多少对数满足a+b = a^b
题解
a+b=a^b其实就是求二进制中每一位都不同的对数。
首先考虑容斥，假设我们知道solve(l,r)就是求[1,l],[1,r]中有多少对答案。
那么最终答案就是solve(r,r)-2solve(l-1,r)+solve(l-1,l-1)
然后这个数位dp，我们正常去跑就行。dp[i][sa][sb]表示考虑第i位，a是否到达的最大值，b是否到达了最大值。然后枚举即可。
代码
#include<bits/stdc++.h>
using namespace std;
long long dp[35][2][2];
long long ans(int l,int r,int x,int sa,int sb){
if(x==-1)return 1;
if(dp[x][sa][sb]!=-1)return dp[x][sa][sb];
int ma=1,mb=1;
if(sa)ma=(l>>x)&1;
if(sb)mb=(r>>x)&1;
dp[x][sa][sb]=0;
for(int i=0;i<=ma;i++){
for(int j=0;j<=mb;j++){
if((i&j)==0){
dp[x][sa][sb]+=ans(l,r,x-1,sa&(i==ma),sb&(j==mb));
}
}
}
return dp[x][sa][sb];
}
long long ans(int l,int r){
if(l<0||r<0)return 0;
memset(dp,-1,sizeof(dp));
return ans(l,r,30,1,1);
}
void solve(){
int l,r;
scanf("%d%d",&l,&r);
cout<<ans(r,r)-2*ans(l-1,r)+ans(l-1,l-1)<<endl;
}
int main(){
int t;
scanf("%d",&t);
while(t--){
solve();
}
}