Atcoder Beginner Contest 115 D Christmas 模拟,递归 B

D - Christmas

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Time limit : 2sec / Memory limit : 1024MB

Score : 400 points

Problem Statement

In some other world, today is Christmas.

Mr. Takaha decides to make a multi-dimensional burger in his party. A level-L burger (L is an integer greater than or equal to 0) is the following thing:

• A level-0 burger is a patty.
• A level-L burger (L≥1) is a bun, a level-(L−1) burger, a patty, another level-(L−1)burger and another bun, stacked vertically in this order from the bottom.

For example, a level-1 burger and a level-2 burger look like BPPPB and BBPPPBPBPPPBB(rotated 90 degrees), where B and P stands for a bun and a patty.

The burger Mr. Takaha will make is a level-N burger. Lunlun the Dachshund will eat X layers from the bottom of this burger (a layer is a patty or a bun). How many patties will she eat?

Constraints

• 1≤N≤50
• 1≤X≤( the total number of layers in a level-N burger )
• N and X are integers.

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Input

Input is given from Standard Input in the following format:

N X

Output

Print the number of patties in the bottom-most X layers from the bottom of a level-N burger.

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Sample Input 1

2 7

Sample Output 1

4

There are 4 patties in the bottom-most 7 layers of a level-2 burger (BBPPPBPBPPPBB).

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Sample Input 2

1 1

Sample Output 2

0

The bottom-most layer of a level-1 burger is a bun.

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Sample Input 3

50 4321098765432109

Sample Output 3

2160549382716056

A level-50 burger is rather thick, to the extent that the number of its layers does not fit into a 32-bit integer.

分析：

比较好玩的一个题目，给你一种关于汉堡的排序方法，然后顺着吃X个问能吃多少个面包，首先看到比较大的数先算了一下会不会爆long long，发现不会之后直接写递归，写完样例没过，检查了半天又把吃一个N等级的所有的个数用一个数组进行存储，中间换了一种形式实现，又找到了一个bug（脑残少算了一个-1），最后AC了。

AC代码：

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <string>
#include <time.h>
#include <queue>
#include <string.h>
#define sf scanf
#define pf printf
#define lf double
#define ll long long
#define p123 printf("123
");
#define pn printf("
");
#define pk printf(" ");
#define p(n) printf("%d",n);
#define pln(n) printf("%d
",n);
#define s(n) scanf("%d",&n);
#define ss(n) scanf("%s",n);
#define ps(n) printf("%s",n);
#define sld(n) scanf("%lld",&n);
#define pld(n) printf("%lld",n);
#define slf(n) scanf("%lf",&n);
#define plf(n) printf("%lf",n);
#define sc(n) scanf("%c",&n);
#define pc(n) printf("%c",n);
#define gc getchar();
#define re(n,a) memset(n,a,sizeof(n));
#define len(a) strlen(a)
#define LL long long
#define eps 1e-6
using namespace std;

ll length[100];
ll sum0[100];
ll sum = 0;
ll f(ll n,ll x0){
    if(n == 1){
        if(x0 == 1){
            return 0;
        }else if(x0 == 2){
            return 1;
        }else if(x0 == 3){
            return 2;
        }else if(x0 == 4){
            return 3;
        }else if(x0 == 5){
            return 3;
        }
    }
    if(x0 == 1){
        return 0;
    }else if(x0 == length[n]){
        return sum0[n];
    }if(x0 == ((length[n]+1)>>1)){
        return sum0[n-1]+1;
    }else if(x0 < ((length[n]+1)>>1)){
        return f(n-1,x0-1);
    }else{
        return f(n-1,(x0-length[n-1]-2))+sum0[n-1]+1;
    }
}


int main() {
    length[0] = 1;
    sum0[0] = 1;
    for(ll i = 1; i <= 50; i ++){
        length[i] = (length[i-1] *2 ) + 3;
        sum0[i] = sum0[i-1]*2+1;
    }
    //pld(sum0[50]); pn
    ll n,x;
    sld(n) sld(x);
    pld(f(n,x)); pn
    return 0;
}
//10 11 12 14 15