hdu 4803 （想法题 卡精度）

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=4803

Poor Warehouse Keeper

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1493    Accepted Submission(s): 428

Problem Description

Jenny is a warehouse keeper. He writes down the entry records everyday. The record is shown on a screen, as follow:

There are only two buttons on the screen. Pressing the button in the first line once increases the number on the first line by 1. The cost per unit remains untouched. For the screen above, after the button in the first line is pressed, the screen will be:

The exact total price is 7.5, but on the screen, only the integral part 7 is shown.
Pressing the button in the second line once increases the number on the second line by 1. The number in the first line remains untouched. For the screen above, after the button in the second line is pressed, the screen will be:

Remember the exact total price is 8.5, but on the screen, only the integral part 8 is shown.
A new record will be like the following:

At that moment, the total price is exact 1.0.
Jenny expects a final screen in form of:

Where x and y are previously given.
What’s the minimal number of pressing of buttons Jenny needs to achieve his goal?

 

Input

There are several (about 50, 000) test cases, please process till EOF.
Each test case contains one line with two integers x(1 <= x <= 10) and y(1 <= y <= 10⁹) separated by a single space - the expected number shown on the screen in the end.

 

Output

For each test case, print the minimal number of pressing of the buttons, or “-1”(without quotes) if there’s no way to achieve his goal.

 

Sample Input

1 13 89 31

 

Sample Output

0511
Hint
For the second test case, one way to achieve is:(1, 1) -> (1, 2) -> (2, 4) -> (2, 5) -> (3, 7.5) -> (3, 8.5)

把上面那个按钮 称为按钮1；

下面那个按钮  称为按钮2；

题意：两种操作，如果按下按钮1，会使数量+1，单价不变，总价=（数量+1）*单价

           如果按下按钮2，会使总价加1，数量不变，单价=(总价+1)/数量;

现在问至少多少步可以使得   按钮1数字为x, 按钮2数字为y;

思路：（1）由题意容易知道至少 需要 x-1步(按上面的按钮)；

            （2）最后可以转化为单价的问题，给定x,y后，不论按哪个按钮，单价只会呈现上身趋势，并且达到最大，那么就由最大的单价倒着推出每两次按下按钮1之间按下按钮2次数的最大值（贪心）

#include <iostream>#include <string.h>#include <string>#include <cstdio>#include <cmath>#include <cstdio>#include <algorithm>typedef long long ll;const double eps=1e-5;using namespace std;int main(){    double x,y;    while(scanf("%lf%lf",&x,&y)!=EOF)    {      if(x>y)      {        cout<<-1<<endl;        continue;      }      double s=(y+1-eps)/x;      int cnt=(int)x-1;      double tmp=1;      for(int i=1;i<=(int)x;i++)      {        int t=(int)(s*i-tmp);        tmp+=t;        cnt+=t;        tmp=tmp/i*(i+1);      }     printf("%d\n",cnt);    }    return 0;}