SCU - 1269 UVA - 10090 Marbles (拓展欧几里得算法的解空间结构)

题目：

Description

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 c2and 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 mi = number of Type 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

这个题目和另外一个题目差不多，我就不解释了。点击打开我的博客
代码：

#include<iostream>
#include<stdio.h>
using namespace std;

long long x, y;

long long gcd(long long  a, long long b)
{
if (a == 0 || b == 0)
{
x = (b == 0);
y = (a == 0);
return a + b;
}
long long r;
if (a < 0)
{
r = gcd(-a, b);
x *= -1;
return r;
}
if (b < 0)
{
r = gcd(a, -b);
y *= -1;
return r;
}
if (a >= b)r = gcd(a%b, b);
else r = gcd(a, b%a);
y -= a / b*x;
x -= b / a*y;
return r;
}

void f(long long n, long long c1, long long n1, long long c2, long long n2)
{
if (c1*n2 < c2*n1)
{
f(n, c2, n2, c1, n1);
long long t = x;
x = y;
y = t;
return;
}
if (n1 == n2)
{
x = -1;
if (n%n2 == 0)
{
x = 0;
y = n / n2;
}
return;
}
long long g = gcd(n1, n2);
if (n % g)
{
x = -1;
return;
}
x *= n / g;
y *= n / g;
long long s = x / (n2 / g);
x -= n2 / g*s;
y += n1 / g*s;
if (x < 0)
{
x += n2 / g;
y -= n1 / g;
}
}

int main()
{
long long n, c1, n1, c2, n2;
while (cin >> n)
{
if (n == 0)break;
cin >> c1 >> n1 >> c2 >> n2;
f(n, c1, n1, c2, n2);
if (x < 0 || y < 0)cout << "failed" << endl;
else cout << x << " " << y << endl;
}
return 0;
}

函数 f 里面加了n1==n2的特判，因为我自己写的这个gcd函数只能计算2个不同的数的gcd