UVA 10090 Marbles（扩展欧几里得）

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 exactlyn₁ marbles

Type 2: each box costs c₂ Taka and can hold exactlyn₂ 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 containsc₁and 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 integersm₁ 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

题意：一个人有n个弹球，现在要把这些弹球全部装进盒子里，第一种盒子每个盒子c1美元，可以恰好装n1个弹球；第二种盒子每个盒子c2元，可以恰好装n2个弹球。找出一种方法把这n个弹球装进盒子，每个盒子都装满，并且花费最少的钱。

分析：假设第一种盒子买m1个，第二种盒子买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，可以解出m1=nx/g, m2=ny/g,所以通解为m1=nx/g + bk/g, m2=ny/g - ak/g,

又因为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。

假设

k1=ceil(-nx/b)

k2=floor(ny/b)

如果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;}