UVA 10090 Marbles（扩展欧几里得）

Posted 2020-09-11 cxchanpin

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 c₁and n₁, and the third line contains c₂ and n₂. Here, c₁, c₂, n₁and n₂ are all positive integers having values smaller than 2,000,000,000.

A test case containing a zero forn 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 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

题意：一个人有n个弹球。如今要把这些弹球所有装进盒子里。第一种盒子每一个盒子c1美元，能够恰好装n1个弹球。另外一种盒子每一个盒子c2元。能够恰好装n2个弹球。找出一种方法把这n个弹球装进盒子，每一个盒子都装满，而且花费最少的钱。

分析：如果第一种盒子买m1个，另外一种盒子买m2个，则n1*m1 + n2*m2 = n。由扩展欧几里得 ax+by=gcd(a,b)= g,如果n%g!=0。则方程无解。

联立两个方程。能够解出m1=nx/g, m2=ny/g,所以通解为m1=nx/g + bk/g, m2=ny/g - ak/g,

又由于m1和m2不能是负数，所以m1>=0, m2>=0,所以k的范围是 -nx/b <= k <= ny/a。且k必须是整数。

如果

k1=ceil(-nx/b)

k2=floor(ny/b)

假设k1>k2的话则k就没有一个可行的解。于是也是无解的情况。

设花费为cost，则cost = c1*m1 + c2*m2,

把m1和m2的表达式代入得

cost=c1*(-xn/g+bk/g)+c2*(yn/g-ak/g) = ((b*c1-a*c2)/g)*k+(c1*x*n+c2*y*n)/g

这是关于k的一次函数。单调性由b*c1-a*c2决定。

若b*c1-a*c2 >= 0，k取最小值（k1）时花费最少；否则，k取最大值（k2）时花费最少。

#include<iostream>
#include<cmath>
using namespace std;
typedef long long LL;

LL extend_gcd(LL a, LL b, LL *x, LL *y)
{
    LL xx, yy, g;
    if(a < b) return extend_gcd(b, a, y, x);
    if(b == 0) {
        *x = 1;
        *y = 0;
        return a;
    }
    else {
        g = extend_gcd(b, a%b, &xx, &yy);
        *x = yy;
        *y = (xx - a/b*yy);
        return g;
    }
}

int main()
{
    LL n, c1, n1, c2, n2, x, y;
    while(cin >> n && n) {
        cin >> c1 >> n1 >> c2 >> n2;
        LL g = extend_gcd(n1, n2, &x, &y);
        if(n % g != 0) {
            cout << "failed" << endl;
            continue;
        }
        LL mink = ceil(-n * x / (double)n2);
        LL maxk = floor(n*y / (double)n1);  // mink <= k <= maxk
        if(mink > maxk) {
            cout << "failed" << endl;
            continue;
        }
        if(c1 * n2 <= c2 * n1) {
            x = n2 / g * maxk + n / g * x;
            y = n / g * y - n1 / g * maxk;
        }
        else {
            x = n2 / g * mink + n / g * x;
            y = n / g * y - n1 / g * mink;
        }
        cout << x << " " << y << endl;
    }
    return 0;
}