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。
1 /* 2 ━━━━━┒ギリギリ♂ eye！ 3 ┓┏┓┏┓┃キリキリ♂ mind！ 4 ┛┗┛┗┛┃＼○／ 5 ┓┏┓┏┓┃ / 6 ┛┗┛┗┛┃ノ) 7 ┓┏┓┏┓┃ 8 ┛┗┛┗┛┃ 9 ┓┏┓┏┓┃ 10 ┛┗┛┗┛┃ 11 ┓┏┓┏┓┃ 12 ┛┗┛┗┛┃ 13 ┓┏┓┏┓┃ 14 ┃┃┃┃┃┃ 15 ┻┻┻┻┻┻ 16 */ 17 #include <algorithm> 18 #include <iostream> 19 #include <iomanip> 20 #include <cstring> 21 #include <climits> 22 #include <complex> 23 #include <fstream> 24 #include <cassert> 25 #include <cstdio> 26 #include <bitset> 27 #include <vector> 28 #include <deque> 29 #include <queue> 30 #include <stack> 31 #include <ctime> 32 #include <set> 33 #include <map> 34 #include <cmath> 35 using namespace std; 36 #define fr first 37 #define sc second 38 #define cl clear 39 #define BUG puts("here!!!") 40 #define W(a) while(a--) 41 #define pb(a) push_back(a) 42 #define Rint(a) scanf("%d", &a) 43 #define Rll(a) scanf("%I64d", &a) 44 #define Rs(a) scanf("%s", a) 45 #define Cin(a) cin >> a 46 #define FRead() freopen("in", "r", stdin) 47 #define FWrite() freopen("out", "w", stdout) 48 #define Rep(i, len) for(int i = 0; i < (len); i++) 49 #define For(i, a, len) for(int i = (a); i < (len); i++) 50 #define Cls(a) memset((a), 0, sizeof(a)) 51 #define Clr(a, x) memset((a), (x), sizeof(a)) 52 #define Fuint(a) memset((a), 0x7f7f, sizeof(a)) 53 #define lrt rt << 1 54 #define rrt rt << 1 | 1 55 #define pi 3.14159265359 56 #define RT return 57 #define lowbit(x) x & (-x) 58 #define onenum(x) __builtin_popcount(x) 59 typedef long long LL; 60 typedef long double LD; 61 typedef unsigned long long Uint; 62 typedef pair<int, int> pii; 63 typedef pair<string, int> psi; 64 typedef map<string, int> msi; 65 typedef vector<int> vi; 66 typedef vector<int> vl; 67 typedef vector<vl> vvl; 68 typedef vector<bool> vb; 69 70 const int maxn = 11; 71 int n, t; 72 double dp[maxn][maxn]; 73 74 75 int main() { 76 // FRead(); 77 Rint(n); Rint(t); 78 int ret = 0; 79 dp[1][1] = t; 80 For(i, 1, n+1) { 81 For(j, 1, i+1) { 82 if(dp[i][j] >= 1) { 83 dp[i][j] -= 1; 84 dp[i+1][j] += (dp[i][j]) / 2; 85 dp[i+1][j+1] += (dp[i][j]) / 2; 86 ret++; 87 } 88 } 89 } 90 printf("%d ", ret); 91 RT 0; 92 }