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

Sample Output 1

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

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Sample Input 2

Sample Output 2

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

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Sample Input 3

Sample Output 3

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\n");
 #define pn printf("\n");
 #define pk printf(" ");
 #define p(n) printf("%d",n);
 #define pln(n) printf("%d\n",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[];
 ll sum0[];
 ll sum = ;
 ll f(ll n,ll x0){
     if(n == ){
         if(x0 == ){
             return ;
         }else if(x0 == ){
             return ;
         }else if(x0 == ){
             return ;
         }else if(x0 == ){
             return ;
         }else if(x0 == ){
             return ;
         }
     }
     if(x0 == ){
         return ;
     }else if(x0 == length[n]){
         return sum0[n];
     }if(x0 == ((length[n]+)>>)){
         return sum0[n-]+;
     }else if(x0 < ((length[n]+)>>)){
         return f(n-,x0-);
     }else{
         return f(n-,(x0-length[n-]-))+sum0[n-]+;
     }
 }

 int main() {
     length[] = ;
     sum0[] = ;
     for(ll i = ; i <= ; i ++){
         length[i] = (length[i-] * ) + ;
         sum0[i] = sum0[i-]*+;
     }
     //pld(sum0[50]); pn
     ll n,x;
     sld(n) sld(x);
     pld(f(n,x)); pn
     return ;
 }
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