华中农业大学第四届程序设计大赛网络同步赛 I

Problem I: Catching Dogs

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 1130  Solved: 292
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Description

 

   Diao Yang keeps many dogs. But today his dogs all run away. Diao Yang has to catch them. To simplify the problem, we assume Diao Yang and all the dogs are on a number axis. The dogs are numbered from 1 to n. At first Diao Yang is on the original point and his speed is v. The i^thdog is on the point a_i and its speed is v_i . Diao Yang will catch the dog by the order of their numbers. Which means only if Diao Yang has caught the i^th dog, he can start to catch the (i+1)^th dog, and immediately. Note that When Diao Yang catching a dog, he will run toward the dog and he will never stop or change his direction until he has caught the dog.Now Diao Yang wants to know how long it takes for him to catch all the dogs.

Input

 

    There are multiple test cases. In each test case, the first line contains two positive integers n(n≤10) and v(1≤v≤10). Then n lines followed, each line contains two integers a_i(|a_i|≤50) and v_i(|v_i|≤5). v_i＜0 means the dog runs toward left and v_i＞0 means the dog runs toward right. The input will end by EOF.

Output

    For each test case, output the answer. The answer should be rounded to 2 digits after decimal point. If Diao Yang cannot catch all the dogs, output “Bad Dog”(without quotes).

Sample Input

2 5
-2 -3
2 3
1 6
2 -2
1 1
0 -1
1 1
-1 -1

Sample Output

6.00
0.25
0.00
Bad Dog

题意：主人抓狗的题目 主人拥有n条狗 主人的速度为v     n条狗编号后 有对应的初始位置ai与初始速度vi vi>0代表方向向右
 当主人抓到第i只狗后才能抓第i+1只狗 如果能抓到所有的狗输出总时间  否则输出 “Bad Dog”；

题解：模拟 注意精度 double 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath> 
using namespace std;
double n,v;
struct node 
{
    double pos;
    double v;
}N[105];
double init=0;
double tim(double target,double vv)
{
    double dis,vvv;
    if(init>target)
    {
     dis=init-target;
     vvv=v+vv;
    }
    else
    {  
     dis=target-init;
     vvv=v-vv;
    }   
    if(dis==0)
    return 0;
    if(vvv<=0)
    return -1;
    return  dis*1.0/vvv*1.0;
}
int main()
{
    while(scanf("%lf %lf",&n,&v)!=EOF)
    {
        memset(N,0,sizeof(N));
        int flag=0;
        for(int i=1;i<=n;i++)
        scanf("%lf %lf",&N[i].pos,&N[i].v); 
        double t=0,tt=0;;
        init=0;
        for(int i=1;i<=n;i++)
        {
          t=tim(N[i].pos+tt*N[i].v,N[i].v);
          if(t<0)
          {
             flag=1;
             break;
          } 
            if(t>0)
            tt+=t;
            init=N[i].pos+N[i].v*tt;
        }
        if(flag)
         cout<<"Bad Dog"<<endl;
        else
         printf("%.2f\n",tt);
    }
    return 0;
}