【Codeforces Round 354 (Div 2)B】【简单模拟】Pyramid of Glasses 漏斗漏水n层t秒灌满多少个

B. Pyramid of Glasses

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream of a beautiful life, so they used champagne glasses to construct a small pyramid. The height of the pyramid is n. The top level consists of only 1 glass, that stands on 2 glasses on the second level (counting from the top), then 3 glasses on the third level and so on.The bottom level consists of n glasses.

Vlad has seen in the movies many times how the champagne beautifully flows from top levels to bottom ones, filling all the glasses simultaneously. So he took a bottle and started to pour it in the glass located at the top of the pyramid.

Each second, Vlad pours to the top glass the amount of champagne equal to the size of exactly one glass. If the glass is already full, but there is some champagne flowing in it, then it pours over the edge of the glass and is equally distributed over two glasses standing under. If the overflowed glass is at the bottom level, then the champagne pours on the table. For the purpose of this problem we consider that champagne is distributed among pyramid glasses immediately. Vlad is interested in the number of completely full glasses if he stops pouring champagne in t seconds.

Pictures below illustrate the pyramid consisting of three levels.

Input

The only line of the input contains two integers n and t (1 ≤ n ≤ 10, 0 ≤ t ≤ 10 000) — the height of the pyramid and the number of seconds Vlad will be pouring champagne from the bottle.

Output

Print the single integer — the number of completely full glasses after t seconds.

Examples

input

3 5

output

4

input

4 8

output

6

Note

In the first sample, the glasses full after 5 seconds are: the top glass, both glasses on the second level and the middle glass at the bottom level. Left and right glasses of the bottom level will be half-empty.

#include<stdio.h>#include<iostream>#include<string.h>#include<string>#include<ctype.h>#include<math.h>#include<set>#include<map>#include<vector>#include<queue>#include<bitset>#include<algorithm>#include<time.h>using namespace std;void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }#define MS(x,y) memset(x,y,sizeof(x))#define MC(x,y) memcpy(x,y,sizeof(x))#define MP(x,y) make_pair(x,y)#define ls o<<1#define rs o<<1|1typedef long long LL;typedef unsigned long long UL;typedef unsigned int UI;template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }const int N = 0, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;int n, t;int a[12][12];int main(){while (~scanf("%d%d", &n, &t)){int one = 1 << n;MS(a, 0); a[1][1] = one * t;int ans = 0;for (int i = 1; i <= n; ++i){for (int j = 1; j <= i; ++j)if (a[i][j] >= one){++ans;a[i + 1][j] += (a[i][j] - one) >> 1;a[i + 1][j + 1] += (a[i][j] - one) >> 1;}}printf("%d\n", ans);}return 0;}/*【trick&&吐槽】暴力大法好，暴力保平安【题意】漏斗漏水n层t秒灌满多少个，具体如图n∈[1,10]t∈[0,1e4]【类型】简单模拟【分析】这题要搞点精致的做法？不如直接暴力模拟！数据太小啦！毕竟水满之后两头各流一半，于是我们给第一层设置2^10（或者2^9）即可，就不必担心精度问题啦。【时间复杂度&&优化】O(n*n)*/