Water Bills

Description


It's 30 century. And water has become very expensive. Recently, your water company raised the rates once more. The table below shows the new rates (consumption is always a positive integer):

Range Price
(cm3) (yuan)
1~100 2
101~10000 3
10001~1000000 5
>1000000 7

This means that, when calculating the amount to pay, the first 100 cm3 have a price of 2 yuan each; the next 9900 cm3 (between 101 and 10000) have a price of 3 Americus each and so on.
For instance, if you use 10123 cm3 water you will have to pay 2 x 100 + 3 x 9900 + 5 x 123 = 30515 yuan.
The evil mathematicians from the company have found a way to gain even more money. Instead of telling you how much energy you have consumed and how much you have to pay, they will show you two numbers related to yourself and to a random neighbor:
A: the total amount to pay if your consumptions were billed together;
B: the absolute value of the difference between the amounts of your bills.
If you can't figure out how much you have to pay, you must pay another 100 yuan for such a "service". You are very economical, and therefore you are sure you cannot possibly consume more than any of your neighbors. So, being smart, you know you can compute how much you have to pay.

Input


The input contains several test cases. Each test case has a single line, containing two integers A and B, separated by a single space, representing the numbers shown to you (1<=A, B<=10^9). You may assume there is always a unique solution, that is, there exists exactly one pair of consumptions that produces those numbers.
The last test case is followed by a line containing two zeros separated by a single space.


Output


For each test case in the input, your program must print a single line containing one integer, representing the amount you have to pay.

Sample Input


1100 300 
35515 27615 
0 0


Sample Output


350 

2900

这个大体思路很明确，就是根据bill捆绑以后的费用计算二者的用水总量，然后在这个范围内找到各自的用水量以满足二者的费用差符合要求即可。一开始做的时候没有考虑时间的问题，用了几乎遍历半数的搜索，导致最后程序用时达260ms之多，分析知可以通过二分查找节约时间，所以讲程序修改为：

#include<stdlib.h>
#include<stdio.h>
int use(int m)//根据输入的水费返回用水量在哪个范围内
{
if(m > 4979900)
return 4;
if(m > 29900)
return 3;
if(m > 200)
return 2;
return 1;
}
int computer(int a)//根据输入的用水量返回水费
{
int price;
price = 0;
if(a > 1000000)
{
price += ((a - 1000000)*7);
a = 1000000;
}
if(a > 10000)
{
price += ((a-10000)*5);
a = 10000;
}
if(a > 100)
{
price += ((a - 100)*3);
a = 100;
}
price += (a * 2);
a = 0;
return price;
}


int main()
{
int m,n,count,level,i,j;
while(scanf("%d %d",&m,&n)&&(m != 0||n !=0))
{
level = use(m);

//计算二者绑定账单后的用水量
switch(level){
case 1:{count = m/2;break;}
case 2:{count = ((m-200)/3)+100;break;}
case 3:{count = ((m-29900)/5)+10000;break;}
case 4:{count = ((m-4979900)/7)+1000000;break;}
default:break;
}

//开始二分查找各自的用水量
i=0;j=count;
while(i<j)//此处无需考虑没有结果的情况
{
if(computer(count - ((i + j)/2))-computer((i+j)/2)>n)
i=((i+j)/2)+1;
else
{
if(computer(count - ((i + j)/2))-computer((i+j)/2)<n)
j=((i+j)/2)-1;
else
break;
}
}
printf("%d\n",computer((i+j)/2));//输出结果
}
return 0;
}

改后用时9ms，分析对程序运行效率的重要性可见一斑。