UVA 10090 - Marbles 拓展欧几里得

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:
T ype 1: each box costs c1 Taka and can hold exactly n1 marbles
T ype 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 mi = number of T ypei 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：给你n个球，给你两种盒子第一种盒子每个盒子c1美元，可以恰好装n1个球；第二种盒子每个盒子c2元，可以恰好装n2个球。找出一种方法把这n个球装进盒子，每个盒子都装满，并且花费最少的钱。

题解：http://blog.csdn.net/lyhvoyage/article/details/37932481

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 <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std ;
typedef long long ll;
const int N=;

/*m1*n1 + m2*n2 = n;

a = n1;
b = n2;

a*x + b*y = g

m1 =  x*n/g + k*b/g;
m2 =  y*n/g - k*a/g;

cost = m1*c1 + m2*c2;

cost = x*n*c1/g + k*b*c1/g  + y*n*c2/g - k*a*c2/g ;

cost = k*(b*c1 - a*c2)/g + (y*n*c2+x*n*c1)/g;
*/
ll ExpGcd(ll a,ll b,ll &x,ll &y)
{
    ll temp,p;
    if(b==)
    {
        x=; y=;
        return a;
    }
    p=ExpGcd(b,a%b,x,y);
    temp=x; x=y; y=temp-(a/b)*y;
    return p;
}
ll n,x,y,c1,c2,n1,n2,l,r;
int main() {
    while(scanf("%lld",&n)!=EOF) {
        if(n == ) break;
        scanf("%lld%lld",&c1,&n1);
        scanf("%lld%lld",&c2,&n2);
        ll g = ExpGcd(n1,n2,x,y);
        if(n % g != ) {
            printf("failed\n");
            continue;
        }
        ll k1 = ceil(-n * x * 1.0/ n2);
        ll k2 = floor(n * y * 1.0/ n1);
        if(k1>k2) {
            printf("failed\n");
            continue;
        }
        if((c2*n1 - c1*n2)>) {
            l = n2 / g * k2 + n/g * x;
            r = n / g * y - n1 / g * k2;
        }
        else {
            l = n2 / g * k1 + n / g * x;
            r = n / g * y - n1 / g * k1;
        }
        printf("%lld %lld\n", l , r);
    }
    return ;
}

代码