uva 10090 Marbles

Problem F

Marbles

Input: standard input

Output: standard output

I have some (say, n) marbles (small glass balls) and I am going to buy some boxes to store them. The boxes are of two types:

Type 1: each box costs c₁ Taka and can hold exactly n₁ marbles

Type 2: each box costs c₂ Taka and can hold exactly n₂ marbles

I want each of the used boxes to be filled to its capacity and also to minimize the total cost of buying them. Since I find it difficult for me to figure out how to distribute my marbles among the boxes, I seek your help. I want your program to be efficient also.

Input

The input file may contain multiple test cases. Each test case begins with a line containing the integer n (1 <= n <= 2,000,000,000). The second line contains c₁ and n₁, and the third line contains c₂ and n₂. Here, c₁, c₂, n₁ and n₂ are all positive integers having values smaller than 2,000,000,000.

A test case containing a zero for n in the first line terminates the input.

Output

For each test case in the input print a line containing the minimum cost solution (two nonnegative integers m₁ and m₂, where m_i = number ofType i boxes required) if one exists, print "failed" otherwise.

If a solution exists, you may assume that it is unique.

 

Sample Input

43
1 3
2 4
40
5 9
5 12
0

 

Sample Output

13 1
failed

___________________________________________________________________

Rezaul Alam Chowdhury

“The easiest way to count cows in a grazing field is to count how many hooves are there and then divide it by four!”

题意很简单：ax+by=c; 求c1x+c2y的最小值。

首先要说一下两个函数的区别。

　　　　floor(1.00001) = 1; floor(1.99999) = 1;

　　　　ceil(1.00001) = 2; ceil(1.99999) =2;

　　　　其实是对函数的取整的问题。

思路：当然，首先要判断是否有解，这个过程。。  g=gcd(a,b);

由于 x = x*c/g + k*(b/g);

y = y*c/g  - k*(a/g);  x>=0 && y>=0 ，因为不能能买负数个东西。

==> x*c/b <=k <=c*y/a;

　　 ok,这个就是k的取值范围。

这里就要用到一个问题，k是整数，如果取值才是合理的呢?

ceil(x*c/b)<=k<=floor(c*y/a);

这里不解释，1.24<=k<=4.25  ==> 2<=k<=4;?? enen .

现在k的范围求出来了，那么现在就是求对应的x，和y的值了。

有式子  c1x+c2y = c1*x+c2*(c-a*x)/b = c1*x - c2*a/b*x + c2*a/b;

就是化简成只有x的情况进行讨论。

我们只需要看c1*x - c2*a/b*x这一部分， x*(  c1-c2*a/b  )

当c1-c2*a/b<0的时候，x应该越到越好，这就可以根据已经求出的k来做了。

当c1-c2*a/b>0的时候，x应该越小越好。同理。

当c1-c2*a/b=0的时候，当然，就随意在前面一种情况里都是一样的。

code:

 #include<iostream>
 #include<stdio.h>
 #include<cstring>
 #include<cstdlib>
 #include<cmath>
 using namespace std;
 typedef long long LL;

 LL Ex_GCD(LL a,LL b,LL &x,LL& y)
 {
     if(b==)
     {
         x=;
         y=;
         return a;
     }
     LL g=Ex_GCD(b,a%b,x,y);
     LL hxl=x-(a/b)*y;
     x=y;
     y=hxl;
     return g;
 }
 int main()
 {
     LL n,c1,n1,c2,n2;
     LL c,a,b,x,y,g;
     while(scanf("%lld",&n)>)
     {
         if(n==)break;
         scanf("%lld%lld",&c1,&n1);
         scanf("%lld%lld",&c2,&n2);
         a=n1;
         b=n2;
         c=n;
         g=Ex_GCD(a,b,x,y);
         if(c%g!=)
         {
             printf("failed\n");
             continue;
         }
         LL lowx =ceil ( -1.0*x*c/(double)b);
         LL upx =  floor(  c*y*1.0/(double)a );
         if(upx<lowx)
         {
             printf("failed\n");
             continue;
         }
         if(c1*b<=a*c2)/** x越大越好，就取上限值 */
         {
             x=x*(c/g)+upx*(b/g);
             y=y*(c/g)-upx*(a/g);
         }
         else
         {
             x=x*(c/g)+lowx*(b/g);
             y=y*(c/g)-lowx*(a/g);
         }
         printf("%lld %lld\n",x,y);
     }
     return ;
 }