巨佬的数学解法

#include <bits/stdc++.h>
using namespace std;
int gcd(int a,int b)
{
    return b==0?a:gcd(b,a%b);
}
int main()
{
    int a,b,c;
    while(cin>>a>>b>>c&&(a&&b&&c))
    {
        a/=gcd(b,c);
        if(a&1)
            cout<<"NO"<<endl;
        else
            cout<<a-1<<endl;
    }
    return 0;
}

蒟蒻的BFS写法

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>

using namespace std;
//多组输入的初始化 
//判断是否需要LL

#define endl '\n'
#define ms(x, y)    memset(x, y, sizeof x)
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PII;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
int a, b, c;
int w[3];
bool st[110][110][110];

struct node 
{
    int v[3];//因为用数组存储便于枚举
    int step;
};


void bfs()
{
    ms(st, false);

    queue<node> q;
    q.push({w[0], 0, 0, 0});
    st[w[0]][0][0] = true;

    while(q.size())
    {
        node t = q.front();
        q.pop();

        if(t.v[0] == t.v[2] && t.v[1] == 0)
        {
            cout << t.step << endl;
            return;
        }

        for(int i = 0; i < 3; i++)
            for(int j = 0; j < 3; j++)
            {
                if(i != j)
                {
                    node temp = t;
                    int minn = min(temp.v[i], w[j] - temp.v[j]);//从i倒水到j
                    temp.v[i] -= minn, temp.v[j] += minn;

                    if(!st[temp.v[0]][temp.v[1]][temp.v[2]])
                    {
                        temp.step ++;
                        q.push(temp);
                        st[temp.v[0]][temp.v[1]][temp.v[2]] = true;
                    }
                }
            }
    }
    cout << "NO" << endl;
}

int main()
{
    while(cin >> w[0] >> w[1] >> w[2])
    {
        if(w[0] == 0 && w[1] == 0 && w[2] == 0)  break;

        if(w[1] > w[2]) swap(w[1], w[2]);

        if(w[0] % 2 == 1)//可乐是奇数    
            cout << "NO" << endl;
        else bfs();
    }
    return 0;
}