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 ba≤y≤b 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 10^51≤Q≤10?5??
• 1 \leq a \leq b \leq 10^{18}1≤a≤b≤10?18??
• 0 \leq x \leq 10^{18}0≤x≤10?18??

Input	Output
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。

#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);
    }
}