标签:else 直接 oid solution new ace 比较 ann 解决
import java.util.Scanner; class test { private double dp[];//用来存储每段的最短时间 private double p[];//用来存储每个充电站点离起点的距离 private int n,c,t; private double vr,vt1,vt2; private double Length; public boolean Solution(){ System.out.print("请输入每个站点离起点的距离:"); Scanner iner=new Scanner(System.in); for(int i=1;i<=this.n;++i){ p[i]=iner.nextInt(); } double curTime=0; for(int i=1;i<=this.n+1;++i){ double minTime=100000.0; for(int j=0;j<i;++j){ if(c>p[i]){ //判断充满电后的距离能不能直接过站点 curTime=p[i]/vt1; }else{ curTime=c/vt1+(p[i]-c)/vt2; //不加油的时间 } //过站点时要分两种情况--一是加油,二是不加油 if(j>0){ //考虑加油的情况 //1.加油时间上,加t秒 //2.提速时间上,减x秒 curTime=dp[i-1]; curTime+=t; if(p[i]-p[i-1] > c){ curTime+=c/vt1+(p[i]-p[i-1]-c)/vt2; } else{ curTime+=(p[i]-p[i-1])/vt1; } } if(curTime<minTime){ //然后再对两者进行比较 minTime=curTime; } } dp[i]=minTime; } int Time= (int) (this.Length / vr); if(dp[this.n+1]<Time){ System.out.println("What a pity rabbit!"); }else{ System.out.println("Good job,rabbit!"); } iner.close(); return true; } public test(int l,int x,int y,int z,int k,int m,int n){ this.Length=l; this.n=x;this.vr=k; this.c=y;this.vt1=m; this.t=z;this.vt2=n; p=new double[this.n+2]; dp=new double[this.n+2]; p[0]=0;p[this.n+1]=l; } } public class rabbit{ public static void main(String[] args) { test space=new test(100,3,5,1,10,20,5); space.Solution(); } }
标签:else 直接 oid solution new ace 比较 ann 解决
原文地址:https://www.cnblogs.com/z2529827226/p/11624553.html