CSU1209-Three Jugs-GCD

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本文介绍了一个有趣的数学问题——三壶取水问题，探讨如何利用三个容量不同的壶通过倒水操作得到任意精确的水量。文章提供了使用扩展欧几里德算法来解决此问题的C++代码实现，并解释了其背后的数学原理。

Y: Three Jugs

Description

We have three jugs A, B, C without any calibration, and an infinite supply of water. There are three types of actions that you can use:
(1) Fill a jug.
(2) Empty a jug.
(3) Pour from one jug to another.
Pouring from one jug to another stops when the first jug is empty or the second jug is full, whichever comes first. For example, if A has 5 gallons, B has 6 gallons and a capacity of 8, then pouring from A to B leaves B full and 3 gallons in A.
Now you need to calculate the minimum accurate gallons of water we can get by using the three jugs.

Input

There is an integer T (1 <= T <= 200) in the first line, means there are T test cases in total.
For each test case, there are three integers a, b, c (1 <= a, b, c <= 10^18) in a line, indicate the capacity (unit: gallon) of the three jugs.

Output

For each test case, you should print one integer in a line, indicates the minimum accurate gallons of water we can get by using the three jugs.

Sample Input

2
3 6 9
6 10 15

Sample Output

3
1

这个题就是找三个数的gcd，用的定理如下：
扩展欧几里德算法是用来在已知a, b求解一组x，y，使它们满足贝祖等式： ax+by = gcd(a, b) =d（解一定存在，根据数论中的相关定理）。扩展欧几里德常用在求解模线性方程及方程组中。
证明过程有需要的说下我再搞

#include <bits/stdc++.h>
#define N 10100
#define INF 0x3f3f3f3f
#define LL long long
#define mem(a,n) memset(a,n,sizeof(a))
#define fread freopen("in.txt","r",stdin)
#define fwrite freopen("out.txt","w",stdout)
using namespace std;
LL gcd(LL a,LL b){
    if(a<b){
        swap(a,b);
    }
    return (b==0?a:gcd(b,a%b));
}
int main()
{
    ios::sync_with_stdio(false);
    LL a,b,c,t;
    cin>>t;
    while(t--){
        cin>>a>>b>>c;
        LL ans=gcd(a,b);
//      cout<<ans<<endl;
        ans=gcd(ans,c);
        cout<<ans<<endl;
    }
    return 0;
}