1 条题解
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0
C :
#include <stdio.h> int main () { double x,y=0,m; scanf("%lf",&m); while(m--) { scanf("%lf",&x); if(x>=0 && x<2) { y=2.5-x; } if(x>=2 && x<4) { y=2-1.5*(x-3)*(x-3); } if(x>=4 && x<6) { y=x/2-1.5; } printf("y=%.1f\n",y); } }
C++ :
#include<iostream> #include<iomanip> using namespace std; int x,m,i; float y; int main() { cin>>m; i=0; while(i<m) { cin>>x; if(x>=0&&x<2) y=-x+2.5; else if(x<4) y=2-1.5*(x-3)*(x-3); else if(x<6) y=x/2-1.5; cout<<"y="<<fixed<<setprecision(1)<<y<<endl; i++; } return 0; }
Pascal :
Program TK1999; var m,x,i:longint; Begin readln(m); for i:=1 to m do begin readln(x); write('y='); if (0<=x)and(x<2) then writeln((-x+2.5):0:1) else if (2<=x)and(x<4) then writeln((2-1.5*(x-3)*(x-3)):0:1) else if (4<=x)and(x<6) then writeln((x/2-1.5):0:1); end; End.
Java :
import java.util.*; public class Main{ public static void main (String[] args) { Scanner in=new Scanner(System.in); double s; int n=in.nextInt(); for(int i=1;i<=n;i++){ double x=in.nextDouble(); if(x>=0&&x<2){ s=2.5-x; System.out.println ("y="+String.format("%.1f",s)); } if(x>=2&&x<4){ s=2-1.5*(x-3)*(x-3); System.out.println ("y="+String.format("%.1f",s)); } if(x>=4&&x<6){ s=x/2-1.5; System.out.println ("y="+String.format("%.1f",s)); } } } }
- 1
信息
- ID
- 1880
- 时间
- 1000ms
- 内存
- 32MiB
- 难度
- (无)
- 标签
- 递交数
- 0
- 已通过
- 0
- 上传者