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 isn. The top level consists of only 1 glass, that stands on 2 glasses on the second level (counting from the top), then 3glasses 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.
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.
Print the single integer — the number of completely full glasses after t seconds.
3 5
4
4 8
6
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.
题意：现在有n层杯子，摆放如图所示，每秒会有1个杯子容量的水倒进最上面的杯子中。如果某一个杯子装满了，那么就会流到下面的杯子，最下面一层的杯子则会流到桌子上，问T秒后有多少个杯子装满了水。
思路：简单模拟下就好了。
#define _CRT_SECURE_NO_DEPRECATE #include <iostream> #include <cstdio> #include <algorithm> #include <cmath> #include <cstring> #include <cstdio> #include <set> using namespace std; typedef long long int LL; const int MAXN = 10 + 5; double cap[MAXN][MAXN]; int vis[MAXN][MAXN],n,t; void dfs(int x, int y, double val) { cap[x][y] += val; if (vis[x][y]) { if (x < 9) { dfs(x + 1, y + 1, cap[x][y] / 2.0); dfs(x + 1, y, cap[x][y] / 2.0); cap[x][y] = 0; } } else { if (cap[x][y] >= 1) { vis[x][y] = 1; cap[x][y] -= 1; if (x < 9) { dfs(x + 1, y + 1, cap[x][y] / 2.0); dfs(x + 1, y, cap[x][y] / 2.0); cap[x][y] = 0; } } } } int main() { //#ifdef CYM // freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); //#endif while (scanf("%d%d", &n,&t)!=EOF) { int ans = 0; memset(cap, 0, sizeof(cap)); memset(vis, 0, sizeof(vis)); dfs(0,0,t); for (int i = 0; i < n; i++) { for (int j = 0; j <= i; j++) { if (vis[i][j]) { ans++; } } } printf("%d ", ans); } return 0; }