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因为第i个人休息j次服从二项分布,算一下组合数。
数据范围小。
求出第i个人休息j次的概率和对应的时间之后,全概率公式暴力统计。
#include<bits/stdc++.h> using namespace std; const int maxn = 101,maxm = 51; int P[maxn],T[maxn],V[maxn]; long long C[maxm][maxm]; double rst[maxn][maxm]; double tim[maxn][maxm]; //const double eps = 1e-11; //#define LOCAL int main() { #ifdef LOCAL freopen("in.txt","r",stdin); #endif int n,m,l; scanf("%d%d%d",&n,&m,&l); C[0][0] = 1; for(int i = 1; i < maxm; i++){ C[i][i] = C[i][0] = 1; for(int j = 1; j < i; j++){ C[i][j] = C[i-1][j] + C[i-1][j-1]; } //cout<<C[i][i/2]<<endl; } for(int i = 0; i < n; i++){ scanf("%d%d%d",P+i,T+i,V+i); } int rt = 0; for(int i = 0; i < n; i++){ double p = P[i]/100., t0 = l*1./V[i]; for(int j = 0; j <= m; j++){ rst[i][j] = C[m][j]*(pow(p,j))*(pow(1-p,m-j)); tim[i][j] = t0+j*T[i]; } //for(int j = 1; j <= m; j++){ // rst[i][j] += rst[i][j-1]; // } } for(int i = 0; i < n; i++){ double ans = 0; for(int j = 0; j <= m; j++){ double t = tim[i][j], wn = 1; for(int k = 0; k < n; k++){ if(k == i) continue; double twn = 0; for(int j2 = m; j2 >= 0; j2--){ if(tim[k][j2] <= t) break; twn += rst[k][j2]; } wn *= twn; } ans += wn*rst[i][j]; } printf("%lf\n",ans); } return 0; }
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原文地址:http://www.cnblogs.com/jerryRey/p/4851281.html