1 条题解
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0
C :
#include <stdio.h> int main(){ int a; scanf("%d",&a); float q; if(a <= 10){ q = 2.5; }else{ q = 2.5 + (a-10) * 1.5; } printf("%.2f",q); return 0; }C++ :
#include<iostream> #include<iomanip> using namespace std; int main(){ double x,i; cin>>i; if(i<=10){ x=2.5; }else if(i>10){ x=2.5+(i-10)*1.5; } cout<<fixed<<setprecision(2)<<x<<endl; }Pascal :
var n:longint; s:real; begin read(n); s:=0; if n<=10 then s:=2.5 else s:=(n-10)*1.5+2.5; writeln(s:0:2); end.Java :
import java.text.DecimalFormat; import java.util.Scanner; public class Main{ public static void main(String[] args) { Scanner scanner=new Scanner(System.in); int w=scanner.nextInt(); float s=0; if(w<=10){ s= (float) 2.5; }else{ s=(float) ((float)(w-10)*1.5+(float)2.5); } DecimalFormat df=new DecimalFormat("#0.00"); System.out.println(df.format(s)); } }Python :
x = int( input() ) if x <= 10: y = 2.5 else: y = 2.5 +(x - 10) * 1.5 print('%.2f' % y)
- 1
信息
- ID
- 42
- 时间
- 1000ms
- 内存
- 512MiB
- 难度
- 7
- 标签
- 递交数
- 21
- 已通过
- 9
- 上传者