hdu 4803 Poor Warehouse Keeper(贪心)

Poor Warehouse Keeper

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5285 Accepted Submission(s): 1356

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 <= 10 ⁹) 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

Sample Output

  
   0
5
11


   
    
     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)

Source

2013ACM/ICPC亚洲区南京站现场赛——题目重现

题意：Number表示个数，prices表示总的价格。一开始Number和prices都是1。每一次按第一个按钮，number的值+1，prices的值为现在的总价格。按第二个按钮，prices的值+1，number的值不变。给出了最终要达到的Number和prices的值。问最少要按几次按钮。

思路：有两种情况，一种是给的prices的值小于number的值，这个时候一定是达不到的。输出-1。
而另一种情况：
要使得单件物品价格最快的接近最终要得到的价格，一定要在number的数量还小的时候按第二个按钮，这样单件价格才能加的最快。所以，当最后将此时单件价格累加起来小于等于y和按下第二个按钮之后累加起来的值仍小于等于y，则这时候可以按下第二个按钮。否则只能按下第一个按钮将单件价格降低。如果按完了第一个按钮，还达不到最终结果那么只能按第二个按钮。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int main()
{
    int x,y;
    while(~scanf("%d%d",&x,&y))
    {
        int t=x-1;
        double per=1;
        double total=1;
        double g=y*1.0/x;
        int cnt=0,now=1,flag=0;
        if(y<x)
            printf("-1\n");
        else
        {
            while(t)
            {
                if(now==x&&(int)total==y)
                {
                    flag=1;
                    break;
                }
                if((int)(total/now*x)==y)
                {
                    cnt+=x-now;
                    flag=1;
                    break;
                }
                if((int)(total/now*x)<=y&&(int)((total+1)/now*x)<=y)//可以按第二个按钮
                {
                    if((int)per<(int)g)//不考虑小数部分
                    {
                        total+=(int)g-(int)per;
                        cnt+=(int)g-(int)per;
                        per=total/now;
                    }
                    else//考虑小数部分
                    {
                        total++;
                        cnt++;
                        per=total/now;
                    }
                }
                else//只能按第一个按钮
                {
                    t--;
                    now++;
                    total+=per;
                    cnt++;
                }
            }
            if(flag==0&&(int)total<y)//now达到了x值但是total没有达到y值，此时就要接着按第二个按钮直到totao达到y值。
            {
                cnt+=y-(int)total;
            }
            printf("%d\n",cnt);
        }
    }
    return 0;
}