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): 5298 Accepted Submission(s): 1363

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 1
3 8
9 31

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)

题意

有一个显示屏上有两个按钮和显示，x代表商品数量，y代表总价(y只显示整数部分)，总价=单价x商品数量，一开始x=y=1(商品单价为1)，我们可以对按钮进行操作，当点击x时，商品数量+1，总价也会随着变化，此时商品单价不发生变化，当点击y时总价+1，此时商品数量保持不变，所以商品单价会增加
题目给你最终的x，y，问最少进行多少次操作能达到x，y

思路

可以利用贪心的思想，每次将单价增加趋近于目标单价，改变x时，单价不变，所以只有在调整y时才能改变单价，那么可以列出基于每次的i(数量)和j(总价)时j能增加的最大值
即j+maxi<y+1x

那么max<(y+1)ix−j
max取比(y+1)ix−j小的整数即可

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#define esp 0.000001using namespace std;int main(){    int x,y;    while(scanf("%d%d",&x,&y)!=EOF)    {        double i,j=1;        double now=1;        int step=0;        int Max;        for(i=1;i<=x;i++,step++)        {                j=now*i;                if(j>=y&&j<y+1) break;            for(;j<y+1;)                {                    Max=(y+1)*i/x-j-esp;                    now=j/i;                    if((j+1)/i*x>=y+1) break;                    j+=Max;                    step+=Max;                    if(j>=y&&j<y+1) goto k;                }        }        k:        if(x>y) printf("-1\n");        else        printf("%d\n",step);    }    return 0;}