[Codeforces676B]Pyramid of Glasses（递推，DP）

题目链接：http://codeforces.com/problemset/problem/676/B

递推，dp(i, j)表示第i层第j个杯子，从第一层开始向下倒，和数塔一样的题。每个杯子1个时间能倒满，从上开始向下倒时间为t，那每个杯子不满的时候对下面的贡献一定是0，所以当dp[i][j]>=1的时候，计数倒满的杯子并且将溢出的酒向下平分。

 /*
 ━━━━━┒ギリギリ♂ eye！
 ┓┏┓┏┓┃キリキリ♂ mind！
 ┛┗┛┗┛┃＼○／
 ┓┏┓┏┓┃ /
 ┛┗┛┗┛┃ノ)
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┃┃┃┃┃┃
 ┻┻┻┻┻┻
 */
 #include <algorithm>
 #include <iostream>
 #include <iomanip>
 #include <cstring>
 #include <climits>
 #include <complex>
 #include <fstream>
 #include <cassert>
 #include <cstdio>
 #include <bitset>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 #include <ctime>
 #include <set>
 #include <map>
 #include <cmath>
 using namespace std;
 #define fr first
 #define sc second
 #define cl clear
 #define BUG puts("here!!!")
 #define W(a) while(a--)
 #define pb(a) push_back(a)
 #define Rint(a) scanf("%d", &a)
 #define Rll(a) scanf("%I64d", &a)
 #define Rs(a) scanf("%s", a)
 #define Cin(a) cin >> a
 #define FRead() freopen("in", "r", stdin)
 #define FWrite() freopen("out", "w", stdout)
 #define Rep(i, len) for(int i = 0; i < (len); i++)
 #define For(i, a, len) for(int i = (a); i < (len); i++)
 #define Cls(a) memset((a), 0, sizeof(a))
 #define Clr(a, x) memset((a), (x), sizeof(a))
 #define Fuint(a) memset((a), 0x7f7f, sizeof(a))
 #define lrt rt << 1
 #define rrt rt << 1 | 1
 #define pi 3.14159265359
 #define RT return
 #define lowbit(x) x & (-x)
 #define onenum(x) __builtin_popcount(x)
 typedef long long LL;
 typedef long double LD;
 typedef unsigned long long Uint;
 typedef pair<int, int> pii;
 typedef pair<string, int> psi;
 typedef map<string, int> msi;
 typedef vector<int> vi;
 typedef vector<int> vl;
 typedef vector<vl> vvl;
 typedef vector<bool> vb;
 
 const int maxn = ;
 int n, t;
 double dp[maxn][maxn];
 
 int main() {
     // FRead();
     Rint(n); Rint(t);
     int ret = ;
     dp[][] = t;
     For(i, , n+) {
         For(j, , i+) {
             if(dp[i][j] >= ) {
                 dp[i][j] -= ;
                 dp[i+][j] += (dp[i][j]) / ;
                 dp[i+][j+] += (dp[i][j]) / ;
                 ret++;
             }
         }
     }
     printf("%d\n", ret);
     RT ;
 }