CodeForces 484A Bits

题目链接：http://codeforces.com/contest/484/problem/A

题目大意：

　　给定一个区间[l, r]，输出这个区间上二进制1的位数最多并且数值最小的数。

分析：

　　典型的位运算

　　需要注意的是，由于数据规模很大，用log2函数会产生超过1e-9以上的误差

代码如下：

#include <bits/stdc++.h>
using namespace std;
 
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define rFor(i,t,s) for (int i = (t); i >= (s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
 
#define LOWBIT(x) ((x)&(-x))
 
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
 
#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,inf,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a))
 
#define pii pair<int,int> 
#define piii pair<pair<int,int>,int> 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
 
inline int gc(){
    static const int BUF = 1e7;
    static char buf[BUF], *bg = buf + BUF, *ed = bg;
     
    if(bg == ed) fread(bg = buf, 1, BUF, stdin);
    return *bg++;
} 
 
inline int ri(){
    int x = 0, f = 1, c = gc();
    for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
    for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
    return x*f;
}
 
typedef long long LL;
typedef unsigned long long uLL;
const double EPS = 1e-9;
const int inf = 1e9 + 9;
const LL mod = 1e9 + 7;
const int maxN = 1e5 + 7;
const LL ONE = 1;
 
int n;
LL l, r, ans;
// lbit ： l的二进制位数 
// rbit ： r的二进制位数 
// hbit ： r的最高二进制位
LL lbit, rbit, hbit;
 
// 二分计算x的二进制位数 
inline int getBits(LL x) {
 
        int cnt = 1;
 
        while((x >> 1) != 0) {
 
            ++cnt;
 
            x >>= 1;
 
        }
 
        return cnt;
 
} 
 
// 取x的最高二进制位 
inline LL HighBit(LL x) {
    while(x & (~LOWBIT(x)) != 0) x &= ~LOWBIT(x);
    return x;
} 
 
int main(){
    cin >> n;
    while(n--) {
        cin >> l >> r;
        ans = 0;
        if(r == 0) {
            cout << ans << endl;
            continue;
        }
        if(l == 0) ++l;
         
        lbit = getBits(l);
        rbit = getBits(r);
        hbit = ONE << (rbit - 1);
         
        // 把相同的高二进制位取下来 
        while(lbit == rbit) {
            l &= hbit - 1;
            r &= hbit - 1;
            ans += hbit;
            if(r == 0) break;
            if(l == 0) ++l;
            lbit = getBits(l);
            rbit = getBits(r);
            hbit = ONE << (rbit - 1);
        }
        if(r == 0) {
            cout << ans << endl;
            continue;
        }
         
        if(getBits(r + 1) > rbit) ans += r; // 如果r是11111，那直接选r 
        else ans += hbit - 1; // 否则选01111 
         
        cout << ans << endl;
    }
    return 0;
}
 
/*
1
7849325874389577 8477432355483974
 
ans:7881299347898367
*/