Codeforces Round #276 (Div. 2)

Codeforces Round #276 (Div. 2)
C. Bitstime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

Let's denote as the number of bits set ("1 'bits) in the binary representation of the non-negative integerX.

You are given multiple queries consisting of pairs of integersLAndR. For each query, findX, Such thatL? ¡U?X? ¡U?R, And is maximum possible. If there are multiple such numbers find the smallest of them.

Input

The first line contains integerNPlease specify the number of queries (1? ¡U?N? ¡U? 10000 ). <symbol · blank "http://www.bkjia.com/kf/ware/vc/" target = "_ blank" class = "keylink"> vcD4KPHA + export + bjwvZW0 + export + aTwvZW0 + LD88ZW0 + cjwvZW0 + export/odw/export/odw /PGVtPnI8L2VtPjxlbT5pPC9lbT4/odw/MTAxOCkuPC9wPgoKCgpPdXRwdXQKPHA + fill = "brush: java; "> 31 22 41 10Output

137

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

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#include 
 
  #include #include 
  
   #include 
   
    using namespace std;__int64  bit[65];int main() {   bit[0]=1;    for(int i = 1; i <= 63; i++) {        bit[i] = 2*bit[i-1];    }    int n;    scanf("%d",&n);    while(n--) {         __int64 l,r;         scanf("%I64d%I64d",&l,&r);         int j;         __int64 sum = 0;         for(j = 0; sum <=r; j++)         {             sum += bit[j];         }         j--;         while(sum > r)         {             sum -= bit[j];             if(sum < l) sum += bit[j];             j--;         }        printf("%I64d\n",sum);    }    return 0;}