Codeforces 1491D（数学+思维）

题意描述

如果\(u\&v=v\)，则表示\(u\)向\(u+v\)连了边，给出\(q\)个询问，每个询问告诉一对\((u,v)\)，问\(u\)和\(v\)是否有边。

思路

如果\(u，显然不存在\(u\&v=v\)这种情况。
如果\(u\)和\(u+v\)之间有边，则\(u+v\)的二进制表示一定是\(u\)的二进制表示某位加上\(1\)，从而导致发生进位操作。所以\(v\)的二进制下的\(1\)的前缀个数一定会多于\(u\)的二进制下的\(1\)的前缀个数，枚举二进制下的位数即可。

AC代码

#include "iostream"
#include "cstring"
#include "string"
#include "vector"
#include "cmath"
#include "algorithm"
#include "map"
#include "set"
#include "queue"
#include "stack"
#include "cassert"
#include "unordered_map"
#include "sstream"
#include "cstdio"

using namespace std;

#define fi first
#define se second
#define PB push_back
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) a.begin(),a.end()
#define rep(x,l,u) for(ll x=l;x=u;x--)
#define sz(x) x.size()
#define IOS ios::sync_with_stdio(false);cin.tie(nullptr);
#define seteps(N) setprecision(N)
#define uni(x) sort(all(x)), x.erase(unique(all(x)), x.end())
#define lson (ind<<1)
#define rson (ind<<1|1) 
#define endl '\n'
#define dbg(x) cerr << #x " = " << (x) << endl
#define mp make_pair

typedef long long ll;
typedef unsigned long long ull;
typedef __int128 lll;
typedef pair PII;
typedef pair PCC;
typedef pair PDD;
typedef pair PLL;
typedef pair PIII;


struct Scanner {
 
    bool hasNext = 1;
    bool hasRead = 1;
 
    int nextInt() {
        hasRead = 0;
        int res = 0;
        char flag = 1, ch = getchar();
        while(ch != EOF && !isdigit(ch)) {
            hasRead = 1;
            flag = (ch == '-') ? -flag : flag;
            ch = getchar();
        }
        while(ch != EOF && isdigit(ch)) {
            hasRead = 1;
            res = res * 10 + (ch - '0');
            ch = getchar();
        }
        if(ch == EOF)
            hasNext = 0;
        return res * flag;
    }
 
    ll nextLL() {
        hasRead = 0;
        ll res = 0;
        char flag = 1, ch = getchar();
        while(ch != EOF && !isdigit(ch)) {
            hasRead = 1;
            flag = (ch == '-') ? -flag : flag;
            ch = getchar();
        }
        while(ch != EOF && isdigit(ch)) {
            hasRead = 1;
            res = res * 10 + (ch - '0');
            ch = getchar();
        }
        if(ch == EOF)
            hasNext = 0;
        return res * flag;
    }
 
    char nextChar() {
        hasRead = 0;
        char ch = getchar();
        while(ch != EOF && isspace(ch)) {
            hasRead = 1;
            ch = getchar();
        }
        if(ch == EOF)
            hasNext = 0;
        return ch;
    }
 
    int nextString(char *str) {
        hasRead = 0;
        int len = 0;
        char ch = getchar();
        while(ch != EOF && isspace(ch)) {
            hasRead = 1;
            ch = getchar();
        }
        while(ch != EOF && !isspace(ch)) {
            hasRead = 1;
            str[++len] = ch;
            ch = getchar();
        }
        str[len + 1] = 0;
        if(ch == EOF)
            hasNext = 0;
        return len;
    }
 
} sc;
 
ll rd() {
    ll x = sc.nextLL();
    return x;
}
 
void rd(int &x) {
    x = sc.nextInt();
}
 
void rd(ll &x) {
    x = sc.nextLL();
}
 
void rd(char &x) {
    x = sc.nextChar();
}
 
void rd(char* x) {
    sc.nextString(x);
}
 
template
void rd(pair &x) {
    rd(x.first);
    rd(x.second);
}
 
template
void rd(T *x, int n) {
    for(int i = 1; i <= n; ++i)
        rd(x[i]);
}

template
void rd(vector &x,int n){
    for(int i = 1; i <= n; ++i)
        rd(x[i]);
}
 
void printInt(int x) {
    if(x < 0) {
        putchar('-');
        x = -x;
    }
    if(x >= 10)
        printInt(x / 10);
    putchar('0' + x % 10);
}
 
void printLL(ll x) {
    if(x < 0) {
        putchar('-');
        x = -x;
    }
    if(x >= 10)
        printLL(x / 10);
    putchar('0' + x % 10);
}
 
void pr(int x, char ch = '\n') {
    printInt(x);
    putchar(ch);
}
 
void pr(ll x, char ch = '\n') {
    printLL(x);
    putchar(ch);
}

//#define LOCAL
template
void pr(pair x, char ch = '\n') {
#ifdef LOCAL
    putchar('<');   
    pr(x.first, ' ');
    pr(x.second, '>');
    putchar(ch);
    return;
#endif //LOCAL
    pr(x.first, ' ');
    pr(x.second, ch);
}
template
void pr(T *x, int n) {
    for(int i = 1; i <= n; ++i)
        pr(x[i], " \n"[i == n]);
}
 
template
void pr(vector &x) {
    int n = x.size();
    for(int i = 1; i <= n - 1; ++i)
        pr(x[i], " \n"[i == n - 1]);
}

const int N=5005;
const int M=1<<12;
const int INF=0x3f3f3f3f;
const int mod=1e9+7;
const lll oone=1;
const double eps=1e-6;
const double pi=acos(-1);

void solve(){
    int q;
    rd(q);
    while(q--){
        int u,v;
        rd(u);rd(v);
        if(u>v) {puts("NO");continue;}
        int cnt=0;
        bool ok=true;
        rep(i,0,30){
            if(u>>i&1) cnt++;
            if(v>>i&1) cnt--;
            if(cnt<0) ok=false;
        }
        if(ok) puts("YES");
        else puts("NO");
    }
}
int main(){
    //IOS;
    //freopen("data.in", "r", stdin);
    //freopen("data.out", "w", stdout);
    //int t;rd(t);
    //rep(i,0,t){
        //printf("Case #%d: ", i+1);
        solve();
    //}
    return 0;
}