Bitwise and Queries

Bitwise and Queriestime limit: Ms
Memory limit: MB

You are givenQQ queries of the formA\ b\ XABX. Count the number of valuesYY such thatA \leq y \leq ba≤ y≤b and < Span class= "Katex-mathml" >x\ \& \ y = x< Span class= "base textstyle uncramped" >x & 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 10^5 1 ≤q≤1 0 ? 5< Span class= "Fontsize-ensurer reset-size5 size5" >?
• 1 \leq a \leq b \leq 10^{18}1 ≤a≤b ≤10 ?18??
• 0 \leq x \leq 10^{18} 0 ≤x≤1 0? 1 8??

Input	Output
41 10 35 10 01 63 732 100 32	26837

X&y==x, explain the x binary representation, there are 1 places, y also have 1, is 0 of the place, Y can be 0 or 1, memory blind a few searches on the line, each can choose 0 or 1, and then judge whether can take 0 or 1.

LR means that at this point y can take any number, as long as the previous x is 0,y is 1 o'clock, recursive selection of 0, it means that the number after Y can be arbitrarily taken, because there are 1 selected 0, then no matter what the choice is not beyond the upper bound.

#include <bits/stdc++.h>using namespaceStd;typedefLong LongLL;Const intN =200005;Const intINF =0x3f3f3f3f;Const intMoD =1000000007; LL dp[ $][2];inta[ $],x[ $]; LL Solve (intCurintLR) {    if(cur = = -)return 1; if(DP[CUR][LR]! =-1)returnDP[CUR][LR]; if(LR) {if(A[cur]) {returnDP[CUR][LR] = solve (cur +1, LR); }        Else {            returnDP[CUR][LR] =2* Solve (cur +1, LR); }    }    Else {        if(A[cur]) {if(X[cur])returnDP[CUR][LR] = solve (cur +1, LR); Else returnDP[CUR][LR] =0; }        Else {            if(X[cur])returnDP[CUR][LR] = Solve (cur+1,1) + Solve (cur+1, LR); Else returnDP[CUR][LR] = Solve (cur+1, LR); }    }}voidPrintint*x) {     for(inti =0; I < -; i++) printf ("%d", X[i]); Puts ("");}intMain () {#ifdef local freopen ("inch","R", stdin);#endif    intT; scanf ("%d",&T);  while(t--) {LL x,y,a; scanf ("%lld%lld%lld",&x,&y,&a); X--;  for(inti =0; I < -; i++) {A[i]= A &1; A>>=1; } Reverse (A,a+ -);//print (A);         for(inti =0; I < -; i++) {X[i]= x &1; X>>=1; } Reverse (X,x+ -);//print (X);memset (dp,-1,sizeofDP); LL Res= Solve (0,0);  for(inti =0; I < -; i++) {X[i]= y &1; Y>>=1; } Reverse (X,x+ -);//print (X);memset (dp,-1,sizeofDP); Res= Solve (0,0) -Res; printf ("%lld\n", RES); }}

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Bitwise and Queries