[POJ2142] The Balance 扩展欧几里得瞎搞

The Balance

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 8292		Accepted: 3620

Description

Ms. Iyo Kiffa-Australis has a balance and only two kinds of weights to measure a dose of medicine. For example, to measure 200mg of aspirin using 300mg weights and 700mg weights, she can put one 700mg weight on the side of the medicine and three 300mg weights on the opposite side (Figure 1). Although she could put four 300mg weights on the medicine side and two 700mg weights on the other (Figure 2), she would not choose this solution because it is less convenient to use more weights.
You are asked to help her by calculating how many weights are required.

Input

The input is a sequence of datasets. A dataset is a line containing three positive integers a, b, and d separated by a space. The following relations hold: a != b, a <= 10000, b <= 10000, and d <= 50000. You may assume that it is possible to measure d mg using a combination of a mg and b mg weights. In other words, you need not consider "no solution" cases.
The end of the input is indicated by a line containing three zeros separated by a space. It is not a dataset.

Output

The output should be composed of lines, each corresponding to an input dataset (a, b, d). An output line should contain two nonnegative integers x and y separated by a space. They should satisfy the following three conditions.

• You can measure dmg using x many amg weights and y many bmg weights.
• The total number of weights (x + y) is the smallest among those pairs of nonnegative integers satisfying the previous condition.
• The total mass of weights (ax + by) is the smallest among those pairs of nonnegative integers satisfying the previous two conditions.

No extra characters (e.g. extra spaces) should appear in the output.

Sample Input

700 300 200
500 200 300
500 200 500
275 110 330
275 110 385
648 375 4002
3 1 10000
0 0 0

Sample Output

1 3
1 1
1 0
0 3
1 1
49 74
3333 1

Source

Japan 2004

 

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题解：就是求Ax + By = C 的一组非负整数解，并且使得x + y 最小；

我们可以通过扩展欧几里得求出x和y的通解；

x = x0 + B * t;

y = y0 - A * t;

|x| + |y| = |x0 + B * t| + |y0 - A * t|；

我们观察上面式子，发现这貌似并不好解；

于是，我们让A > B， 观察发现x的上升趋势比y的下降趋势小， 而我们要求的是绝对值的最小， 所以对上面那个式子影响最大的其实是y；

我们只需要让y等于0，然后在这周围寻找答案就行了；

有些玄学；

 

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

inline int gcd(int a, int b){return b == 0 ? a : gcd(b, a % b);}
inline void exgcd(int a, int b, int &x, int &y)
{
    if (b == 0) {x = 1, y = 0; return ;}
    exgcd(b, a % b, x, y);
    int tmp = x;
    x = y;
    y = tmp - a / b * y;
}

signed main()
{
    int A, B, C;
    while (scanf("%d%d%d", &A ,&B, &C) != EOF)
    {
        if (!A and !B and !C) break;
        bool flag = 0;
        if (A < B) swap(A, B), flag = 1;
        int x0, y0;
        int g = gcd(A, B);
        if (C % g != 0) {puts("laji");continue;}
        exgcd(A, B, x0, y0);
        A /= g, B /= g, C /= g;
        x0 = x0 * C;
        y0 = y0 * C;
        int t = y0 / A;
        int x = 1e9, y=1e9;
        for (register int i = t - 5 ; i <= t + 5 ; i ++)
        {
            int tx = x0 + B * i;
            int ty = y0 - A * i;
            if (abs(tx) + abs(ty) < x + y ) x = abs(tx), y = abs(ty);
        }
        
        if (!flag) printf("%d %d\n", x, y);
        if (flag) printf("%d %d\n", y, x);
    }
    return 0;
}