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Bitwise And Queries
Bitwise And Queries
Time limit: 1500 ms
Memory limit: 128 MBYou 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
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); }}
Bitwise And Queries
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