数学专项counting:UVa 10081
2013-03-28 22:59
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递推关系很容易得出,而且数据最大为10^100,所以用double类型储存数据即可,无需使用高精度。
#include <iostream> #include <fstream> #include <cstring> #include <cstdio> #include <cmath> using namespace std; int n,k; double dp[110][12]; int vis[110][12]; double dfs(int dep,int i) { if(vis[dep][i]) return dp[dep][i]; if(dep==n) return dp[dep][i]=1; vis[dep][i]=1; double &ans=dp[dep][i]; ans=dfs(dep+1,i); if(i>0) ans+=dfs(dep+1,i-1); if(i<k) ans+=dfs(dep+1,i+1); return ans; } int main() { freopen("in.txt","r",stdin); while(cin>>k>>n) { memset(vis,0,sizeof(vis)); double ans=0; for(int i=0;i<=k;i++) ans+=dfs(1,i); printf("%.5f\n",ans*100/pow((k+1)*1.0,n)); } return 0; }
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