[数据结构与算法]abc238 D (签的并不是很到)

题目
题意:
x & y = a
x + y = s
判断是否存在非负整数x、y满足条件
思路: 首先考虑到两个数至少都是a,s >= 2 *a.
先假定x和y都是a，然后去凑s-2 * a剩余位置。如果剩余位置在a中出现过，那寄了，因为这一位用过。
s >= 2 * a and ( (s-2 *a) & a) == 0
或者特殊一点，x = a,y = s-a. x & y == a
时间复杂度: O(1)
代码:

// Problem: D - AND and SUM
// Contest: AtCoder - Monoxer Programming Contest 2022（AtCoder Beginner Contest 238）
// URL: https://atcoder.jp/contests/abc238/tasks/abc238_d
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<complex>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<unordered_map>
#include<list>
#include<set>
#include<queue>
#include<stack>
#define OldTomato ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr)
#define fir(i,a,b) for(int i=a;i<=b;++i) 
#define mem(a,x) memset(a,x,sizeof(a))
#define p_ priority_queue
// round() 四舍五入 ceil() 向上取整 floor() 向下取整
// lower_bound(a.begin(),a.end(),tmp,greater<ll>()) 第一个小于等于的
// #define int long long //QAQ
using namespace std;
typedef complex<double> CP;
typedef pair<int,int> PII;
typedef long long ll;
// typedef __int128 it;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const ll inf = 1e18;
const int N = 2e5+10;
const int M = 1e6+10;
const int mod = 1e9+7;
const double eps = 1e-6;
inline int lowbit(int x){ return x&(-x);}
template<typename T>void write(T x)
{
    if(x<0)
    {
        putchar('-');
        x=-x;
    }
    if(x>9)
    {
        write(x/10);
    }
    putchar(x%10+'0');
}
template<typename T> void read(T &x)
{
    x = 0;char ch = getchar();ll f = 1;
    while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}
    while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}
int n,m,k,T;
void solve()
{
   ll a,s; read(a),read(s);
   if(s >= 2*a && ( ((s-2*a)&a) == 0) ) puts("Yes");
   else puts("No");
   
   // if(((s-a)&a) == a) puts("Yes");
   // else puts("No");
}
signed main(void)
{  
   // T = 1;
   // OldTomato; cin>>T;
   read(T);
   while(T--)
   {
   	 solve();
   }
   return 0;
}