1 solutions
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0
C :
#include <stdio.h> int main() { int a; float b,p1=0.4,p2=0.8,p3=0.9,p4=0.7,p=0.85; while(scanf("%d",&a) != EOF) { if (a==1) b=p1*p2*p3+p1*p4-p1*p2*p3*p4; else if(a==2) b=2*p*p+2*p*p*p-5*p*p*p*p+2*p*p*p*p*p; else if (a==0) b=0.000001; printf("%.6f\n",b); } return 0; }C++ :
#include<string> int main() { int a; float p1=0.4; float p2=0.8; float p3=0.9; float p4=0.7; float p=0.85; while(scanf("%d",&a) != EOF) {if (a==1){ printf("%.6f\n",p1*p2*p3+p1*p4-p1*p2*p3*p4);} else if(a==2) {printf("%.6f\n",2*p*p+2*p*p*p-5*p*p*p*p+2*p*p*p*p*p);} else { printf("%.6f\n",0.000001); } } }Java :
import java.text.DecimalFormat; import java.util.Scanner; public class Main { public static void main(String[] args){ float f[] = new float[2]; Scanner reader = new Scanner(System.in); int a; double p =0.85 , p1 =0.4 , p2 = 0.8, p3 = 0.9, p4 = 0.7; DecimalFormat df=new DecimalFormat("#0.000000"); while(reader.hasNext()){ a = reader.nextInt(); if(a==1){ f = null; f =new float[1]; f[0]=(float)(p1*p4+p1*p2*p3-p1*p2*p3*p4); } else if(a==2){ f = null; f =new float[1]; f[0]=(float)(2*p*p*p*p*p-5*p*p*p*p+2*p*p*p+2*p*p); } else { f = null; f =new float[1]; f[0] = (float) 0.000001; } /*switch(a){ case 1:f = null; f =new float[1]; f[0]=(float)28/45; break; case 2:f = null; f =new float[1]; f[0]=(float)1/45; break; case 3:f = null; f =new float[1]; f[0]=(float)16/45; break; case 4:f = null; f =new float[1]; f[0]=(float)1/5; break; default: f = null; f =new float[1]; f[0]=(float)0.0001; break; }*/ for(int i=0;i<f.length;i++){ System.out.println(df.format(f[i])); } } } }
- 1
Information
- ID
- 16431
- Time
- 1000ms
- Memory
- 128MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
- 0
- Uploaded By