Bits
Let's denote as the number of bits set ('1' bits) in the binary representation of the non-negative integer x.
You are given multiple queries consisting of pairs of integers l and r. For each query, find the x, such that l ≤ x ≤ r, and is maximum possible. If there are multiple such numbers find the smallest of them.
Input
The first line contains integer n — the number of queries (1 ≤ n ≤ 10000).
Each of the following n lines contain two integers li, ri — the arguments for the corresponding query (0 ≤ li ≤ ri ≤ 1018).
Output
For each query print the answer in a separate line.
Examples
3
1 2
2 4
1 10
1
3
7
Note
The binary representations of numbers from 1 to 10 are listed below:
110 = 12
210 = 102
310 = 112
410 = 1002
510 = 1012
610 = 1102
710 = 1112
810 = 10002
910 = 10012
1010 = 10102
sol:因为二进制一些奇奇怪怪的东西,我们可以得到以下的乱搞,先一位位加上去加到比R大了为止,然后要减到恰好大于等于L,所以尽可能减大的,并没有减过的位
#include <bits/stdc++.h> using namespace std; typedef long long ll; inline ll read() { ll s=0; bool f=0; char ch=' '; while(!isdigit(ch)) { f|=(ch=='-'); ch=getchar(); } while(isdigit(ch)) { s=(s<<3)+(s<<1)+(ch^48); ch=getchar(); } return (f)?(-s):(s); } #define R(x) x=read() inline void write(ll x) { if(x<0) { putchar('-'); x=-x; } if(x<10) { putchar(x+'0'); return; } write(x/10); putchar((x%10)+'0'); return; } #define W(x) write(x),putchar(' ') #define Wl(x) write(x),putchar(' ') int Q; bool Arr[65]; //const ll a[65]={0,1,3,7,15,31,63,127,255,511,1023,2047,4095,8191,16383,32767,65535,131071,262143,524287,1048575,2097151,4194303,8388607,16777215,33554431,67108863,134217727,268435455,536870911,1073741823,2147483647,4294967295,8589934591,17179869183,34359738367,68719476735,137438953471,274877906943,549755813887,1099511627775,2199023255551,4398046511103,8796093022207,17592186044415,35184372088831,70368744177663,140737488355327,281474976710655,562949953421311,1125899906842623,2251799813685247,4503599627370495,9007199254740991,18014398509481983,36028797018963967,72057594037927935,144115188075855871,288230376151711743,576460752303423487,1152921504606846975,2305843009213693951}; int main() { int i,j;; R(Q); while(Q--) { ll L=read(),R=read(),ans=0,oo=-1; for(i=0;ans<R;i++) { ++oo; ans+=(1ll<<(oo)); } memset(Arr,0,sizeof Arr); while(ans>R) { for(j=oo;~j;j--) if(!Arr[j]) { if(ans-(1ll<<j)>=L) { ans-=(1ll<<j); Arr[j]=1; break; } } } Wl(ans); } return 0; } /* Input 3 1 2 2 4 1 10 Output 1 3 7 */