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Published on 2017-09-02 11:31 in 暂未分类 with qscqesze

# Codeforces Round #597 (Div. 2) F. Daniel and Spring Cleaning 数位dp

## 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.

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).

## 题解

a+b=a^b其实就是求二进制中每一位都不同的对数。

## 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();
}
}
posted @ 2019-11-04 01:52  qscqesze  阅读(236)  评论(0编辑  收藏