E - Pyramid of Glasses

模拟

#include<stdio.h>#include<string.h>double dp[15][15];int n,t;void solve(){    dp[1][1]+=1;    for(int i=1;i<=n;i++)    {        for(int j=1;j<=i;j++)        {            if(dp[i][j]<=1)                continue;            double p=dp[i][j]-1;            dp[i][j]=1;            dp[i+1][j]+=p/2;            dp[i+1][j+1]+=p/2;        }    }}int main(){    int ans=0;    scanf("%d%d",&n,&t);    memset(dp,0,sizeof(dp));    for(int k=1;k<=t;k++)        solve();    for(int i=1;i<=n;i++)    {        for(int j=1;j<=i;j++)        {            if(dp[i][j]>=1)                ans++;        }    }    printf("%d\n",ans);    return 0;}