九度OJ 1062 分段函数

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
double y(double x){
    double result;
    if(0<=x&&x<2.0)
    {
        result=-1*x+2.5;
    }
    else if(2<=x&&x<4){
        result=2-1.5*(x-3.0)*(x-3.0);
    }
    else{
        result=x/2-1.5;
    }
    return result;
}
int main(int argc, char *argv[])
{
    double N;
    while(~scanf("%lf",&N))
    {
        printf("%.3lf\n",y(N));
    }
    return 0;
}
 
/**************************************************************
    Problem: 1062
    User: kirchhoff
    Language: C
    Result: Accepted
    Time:10 ms
    Memory:912 kb
****************************************************************/