1 条题解
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0
C :
#include <stdio.h> #include <math.h> int main() { int a; double answer[5]; double p[3]; double E[4]; //E1,2,3分别对应x1,2,3的期望 E[1]=5*(1.0/2); E[2]=E[3]=6*(1.0/3); while(scanf("%d",&a) != EOF) { if(a==1) { p[0]=10*pow(0.5,5); p[1]=15*pow(1.0/3,2)*pow(2.0/3,4); p[2]=6*pow(1.0/3,5)*(2.0/3); answer[1]=p[0]*p[1]*p[2]; printf("%0.6f\n",answer[1]);//P{X1=2,X2=2,X3=5}, } else if(a==2) { answer[2]=E[1]*E[2]*E[3]; printf("%0.6f\n",answer[2]);//E(X1,X2,X3) } else if(a==3) { answer[3]=E[1]-E[2]; printf("%0.6f\n",answer[3]);//E(X1-X2) } else if(a==4) { answer[4]=E[1]-2*E[2]; printf("%0.6f\n",answer[4]);//E(X1-2*X2) } else if(a==0) { answer[0] = 0.000001; printf("%0.6f\n",answer[0]); } } return 0; }
C++ :
#include <iostream> using namespace std; int main() { int N; while(scanf("%d",&N) != EOF) { if (N==1) cout<<"0.001694"<<endl; else if(N==2) cout<<"10.000000"<<endl; else if(N==3) cout<<"0.500000"<<endl; else if(N==4) cout<<"-1.500000"<<endl; else cout<<"0.000001"<<endl; } return 0; }
Java :
import java.util.*; public class Main { public static void main(String args[]) { Scanner cin = new Scanner(System.in); int a; double b; while (cin.hasNext()) { a = cin.nextInt(); if (a==1){ b=0.001694; System.out.printf("%.6f\n",b); } else if (a==2){ b=10.000000; System.out.printf("%.6f\n",b);} else if (a==3){ b=0.500000; System.out.printf("%.6f\n",b);} else if (a==4){ b=-1.500000; System.out.printf("%.6f\n",b);} else if (a==0){ b=0.000001; System.out.printf("%.6f\n",b);} } } }
- 1
信息
- ID
- 865
- 时间
- 1000ms
- 内存
- 128MiB
- 难度
- (无)
- 标签
- 递交数
- 0
- 已通过
- 0
- 上传者