[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。

 /*

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 */

 #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 ;

 }