题意:求用N(1<=N<=100)个骰子掷出M(1<=M<=600)的概率
分析:直接求概率可能出现6^100次方,会爆精度。可以用一个数组dp[i][j]记录用i个骰子掷出j的概率。i为0时无论j是多少,概率都是0。i为1时,j从1-6的概率都是1/6。其余可以递推得到
dp[i][j] = 0 (j<i || j>6*i),sigma(dp[i-1][max(0,j-k)]) (1<=k<=6)
#include<stdio.h> #include<cstring> #include<algorithm> #include<iostream> #include<cmath> #include<vector> #include<map> #include<unordered_map> using namespace std; typedef long long LL; const int maxn = 1e4+5; const int INF= 0x3f3f3f3F; const int mod = 200907; double dp[105][605]; void pre() { for(int i=1;i<=6;++i) dp[1][i]=(double)(1.0/6); for(int i=2;i<=100;++i){ for(int j=1;j<=600;++j){ if(j<i) dp[i][j] = 0.0; else{ dp[i][j] =0; for(int k=max(1,j-6);k<j;++k){ dp[i][j] += 1.0*dp[i-1][k]/6.0; } } } } } int main() { #ifndef ONLINE_JUDGE freopen("in.txt","r",stdin); freopen("out.txt","w",stdout); #endif dp[0][0]=0; pre(); int T,N,M,Q,u,v,tmp,K; while(scanf("%d%d",&N,&M)==2){ printf("%.2f\n",100.0*dp[N][M]); } return 0; }
原文地址:https://www.cnblogs.com/xiuwenli/p/9424857.html
时间: 2024-10-10 01:24:37