2015
09-18

# Poor Warehouse Keeper

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?

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.

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.

1 1
3 8
9 31

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)


# 题目链接：hdu 4803 Poor Warehouse Keeper

y+1>txi

t<(y+1)ix

    即，每次对于一个数量，尽量加总价，使得单价尽量大，并且保证在和面加数量时不会大于上限，因为单价大的话，加一次数量总价接近目标值的速度会更快。

#include <cstdio>
#include <cstring>
#include <cmath>

const double eps = 1e-9;

int main () {
double x, y;

while (scanf("%lf%lf", &x, &y) == 2) {

if (x > y) {
printf("-1\n");
continue;
}

double k = (y+1-eps) / x;
int cnt = (int)x - 1;

double tmp = 1;
for (int i = 1; i <= (int)x; i++) {
double t = i * k;
int u = (int)(t-tmp);
tmp += u;

tmp = tmp * (i+1) / i;
cnt += u;
}
printf("%d\n", cnt);
}
return 0;
}