CSU1209-Three Jugs-GCD

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

23 6 96 10 15

Sample Output

31

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;}