Bitwise And Queries

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Bitwise And Queries

Time limit: 1500 ms
Memory limit: 128 MB

You are given QQ queries of the form a\ b\ xa b x. Count the number of values yy such that a \leq y \leq bayb and x\ \& \ y = xx & y=x, where we denote by \&& the bitwise and operation.

Standard input

The first line contains a single integer QQ.

Each of the following QQ lines contains three integers a\ b\ xa b x, representing a query.

Standard output

Output QQ lines, each containing a single integer representing the answer to a query.

Constraints and notes

  • 1 \leq Q \leq 10^51Q10?5??
  • 1 \leq a \leq b \leq 10^{18}1ab10?18??
  • 0 \leq x \leq 10^{18}0x10?18??
InputOutput
4
1 10 3
5 10 0
1 63 7
32 100 32
2
6
8
37
 

 

x&y==x,说明x二进制表示上,有1的地方,y也要有1,是0的地方,y可以是0或者1,记忆化瞎几把搜一下就行了,每次可以选择的有0或1,再判断一下能否取0或1。

lr表示此时这个位置y能否取任意数,只要前面x为0,y为1时,递归选0的话,就说明y后面的数可以任意取了,因为前面有1选了0,那后面无论选什么都不可能超过上界了。
技术分享
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 200005;
const int inf = 0x3f3f3f3f;
const int mod = 1000000007;
LL dp[65][2];
int A[65],X[65];
LL solve(int cur,int lr){
    if(cur == 64)return 1;
    if(dp[cur][lr] != -1)return dp[cur][lr];
    if(lr){
        if(A[cur]){
            return dp[cur][lr] = solve(cur + 1,lr);
        }
        else {
            return dp[cur][lr] = 2 * solve(cur + 1,lr);
        }
    }
    else {
        if(A[cur]){
            if(X[cur])return dp[cur][lr] = solve(cur + 1,lr);
            else return dp[cur][lr] = 0;
        }
        else {
            if(X[cur])return dp[cur][lr] = solve(cur+1,1) + solve(cur+1,lr);
            else return dp[cur][lr] = solve(cur+1,lr);
        }
    }
}
void print(int *x){
    for(int i = 0; i < 64; i++)
        printf("%d",x[i]);
    puts("");
}
int main(){
#ifdef local
    freopen("in", "r", stdin);
#endif
    int T;
    scanf("%d",&T);
    while(T--){
        LL x,y,a;
        scanf("%lld%lld%lld",&x,&y,&a);
        x--;
        for(int i = 0; i < 64; i++){
            A[i] = a & 1;
            a >>= 1;
        }
        reverse(A,A+64);
//        print(A);
        for(int i = 0; i < 64; i++){
            X[i] = x & 1;
            x >>= 1;
        }
        reverse(X,X+64);
//        print(X);
        memset(dp,-1,sizeof dp);
        LL res = solve(0,0);
        for(int i = 0; i < 64; i++){
            X[i] = y & 1;
            y >>= 1;
        }
        reverse(X,X+64);
//        print(X);
        memset(dp,-1,sizeof dp);
        res = solve(0,0) - res;
        printf("%lld\n",res);
    }
}
View Code

 

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