codeforces 676B B. Pyramid of Glasses(模拟)

题目链接：[b]B. Pyramid of Glasses[/b]time limit per test
1 secondmemory limit per test
256 megabytesinput
standard inputoutput
standard outputMary 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.InputThe 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.OutputPrint the single integer — the number of completely full glasses after t seconds.Examplesinput

3 5

output

4

input

4 8

output

6

题意：

像图中那样摆放杯子,共n层,问t分之后有多少杯子被倒满了;

思路：

模拟倒酒的过程,然后再统计就好了;

ＡＣ代码：

#include <bits/stdc++.h>
/*
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <cstring>
#include <algorithm>
#include <cstdio>
*/
using namespace std;
#define Riep(n) for(int i=1;i<=n;i++)
#define Riop(n) for(int i=0;i<n;i++)
#define Rjep(n) for(int j=1;j<=n;j++)
#define Rjop(n) for(int j=0;j<n;j++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
const LL mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=1e6+4;
int n,t;
double a[20][20];
void fun()
{
a[1][1]+=1;
Riep(n)
{
Rjep(i)
{
if(a[i][j]>1)
{
a[i+1][j]+=(a[i][j]-1)/2;
a[i+1][j+1]+=(a[i][j]-1)/2;
a[i][j]=1;
}
}
}
}
int main()
{
scanf("%d%d",&n,&t);
for(int i=1;i<=t;i++)fun();
int ans=0;
Riep(n)
{
Rjep(i)if(a[i][j]>=1)ans++;
}
printf("%d\n",ans);

return 0;
}