zstu19一月月赛 duxing201606的原味鸡树

 duxing201606的原味鸡树

题意：

　　给定一颗有n（n<=1e9）个节点的完全二叉树，1e5次询问，问某个节点有几个子节点。

思路：

　　　自己在月赛上没有思路，问了zfq才知道。

　　　设两个指标，L、R，因为是范围，所以每次L向左孩子一直下去，R向右孩子一直下去,每次下探答案就要加上2的i次，L～R间就是根节点所表示的范围。当n出了L，R区间，退出。

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>

using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "
";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '
'

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
#define max3(a,b,c) max(max(a,b), c);
#define min3(a,b,c) min(min(a,b), c);
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 9999973;
const double esp = 1e-8;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黄金分割点
const double tPHI=0.38196601;


template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}
/*-----------------------showtime----------------------*/


int main(){
            int n;int m;
            scanf("%d%d", &n, &m);
            for(int i=1; i<=m; i++){
                int x;  scanf("%d", &x);
                int l=x,r=x;
                ll res = 0,now = 1;
                for(;;){
                    l = l*2;
                    r = r*2+1;
                    res = res + now;
                    now = now * 2;
                    if( l<= n && n<r){
                        res += n - l + 1;
                        break;
                    }
                    if( n < l) break;
                }
                printf("%lld
", res);
            }

            return 0;
}