AIZU2303marathon match 概率

来源：互联网发布：windows程序设计第7版编辑：程序博客网时间：2024/06/17 16:12
#include <cstdio>#include <iostream>#include <cstring>#include <cmath>using namespace std;#define LL long longdouble p[200];int t[200];int v[200];LL C[100][100];double dp[105][55];int main(){    int n,m,l;    cin >>  n >> m >> l;    for(int i = 0;i < n;i++){        int x,tt,vv;        scanf("%d%d%d",&x,&tt,&vv);        p[i] = x / 100.0,t[i] = tt,v[i] = vv;    }    for(int i = 0;i <= 50;i++){        C[i][0] = 1;        C[i][i] = 1;        for(int j = 1;j < i;j++){            C[i][j] = C[i-1][j]+C[i-1][j-1];            //cout<<C[i][j] << endl;        }    }    for(int i = 0;i < n;i++){        dp[i][m+1] = 0;        for(int j = m;j >= 0;j--){ //休息>j次            dp[i][j] = dp[i][j+1] + C[m][j]*pow(p[i],j)*pow(1-p[i],m-j);            //cout << dp[i][j] << endl;        }    }    for(int i = 0;i < n;i++){        double ans = 0;        if(v[i] != 0){            for(int j = 0;j <= m;j++){                double ut = l*1.0/v[i] + j*t[i];                double pa = C[m][j]*pow(p[i],j)*pow(1-p[i],m-j);                for(int k = 0;k < n;k++){//赢k个人的概率                    if(k != i){                        if(v[k] == 0){                            pa *= 1;                            continue;                        }                        int tim;                        if(t[k] == 0){                            if(l*1.0/v[k] <= ut){                                pa *= 0;                            }                            else{                                pa *= 1;                            }                            continue;                        }                        tim = floor((ut-(l*1.0/v[k]))/t[k]);                        while((tim*t[k]+l*1.0/v[k]) <= ut) tim++;                        if(tim > m){                            pa *= 0;                        }                        else if(tim < 0){                            pa *= 1;                        }else{                            pa *= dp[k][tim];                        }                    }                }                //printf("%lf\n",pa);                ans += pa;            }        }        printf("%.8f\n",ans);    }    return 0;}
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