【codeforces】 Round #287 (Div. 2) C. Guess Your Way Out!-优快云博客

本文介绍了一种解决数学问题的高效算法，通过使用多种数据结构如队列、堆栈、映射等，实现了快速求解。算法核心在于对输入进行预处理，利用算法优化减少计算复杂度。

数学题。。YY两下就可以了。。

#include <iostream>
#include <queue> 
#include <stack> 
#include <map> 
#include <set> 
#include <bitset> 
#include <cstdio> 
#include <algorithm> 
#include <cstring> 
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 200005
#define maxm 400005
#define eps 1e-10
#define mod 1000000007
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid 
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
//#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
// head

LL n, h;
LL two[100];
LL a[100];

void read()
{
	scanf("%I64d%I64d", &h, &n);
}

void work()
{
	LL cnt = 0, t = 0, ans = 0;
	two[0] = 1;
	for(int i = 1; i <= h; i++) two[i] = two[i-1] * 2;
	for(int i = 0; i < h; i++) t += two[i];
	for(int i = 1; i <= h; i++) two[i] += two[i-1];
	t += n;
	while(t) {
		a[cnt++] = t % 2;
		t /= 2;
	}
	LL now = 0;
	for(int i = cnt - 2, j = h - 1; i >= 0; i--, j--) {
		ans++;
		if(a[i] != now) ans += two[j], now ^= 1;
		now ^= 1;
	}
	printf("%I64d\n", ans);
}

int main(void)
{
	read();
	work();

	return 0;
}