扩展欧几里德

Ax+By = c。当且仅当c为gcd(A, B)的倍数时有解。

扩展gcd，不仅求出gcd而且求出解x,y。常用于求解ax≡1(mod m) => ax+by=1。

递归实现，结束状态时x=1，y=0。更新x和y：x=y1, y = x1-(a/b)*y1。

51Nod1352 https://vjudge.net/contest/218366#problem/C

中间溢出要用long long

Ax+By=N-1参考：https://blog.csdn.net/Viscu/article/details/52735003

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<vector>
#include<stack>
#include<queue>
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define IO ios::sync_with_stdio(false);cin.tie(0);
#define INF 0x3f3f3f3f
#define MAXN 100010
const int MOD=1e9+7;
typedef long long ll;
using namespace std;
ll n, a, b, t, x, y, cnt;
ll ext_gcd(ll a, ll b, ll &x, ll &y)
{
    if(b == 0){
        x = 1;
        y = 0;
        return a;
    }
    ll ans = ext_gcd(b, a%b, x, y);
    ll t = x;
    x = y;
    y = t-a/b*y;
    return ans;
}
int main()
{
    cin >> t;
    while(t--){
        cnt=0;
        cin >> n >> a >> b;
        ll r = ext_gcd(a, b, x, y); 
        if((n+1)%r!=0){//无解 
            cout << "0" << endl; 
        } 
        else{
            ll mod = b/r, mini, lcm = a*b/r;//lcm = a*b/gcd, 再除a就是mod 
            mini = (x*(n+1)/r%mod+mod)%mod;//mini为最小非负x 
            if(mini == 0){//0不能为除数，且给定集合不含0 
                mini = mod;
            }
            if(mini*a>n){//如果一开始就大了 
                cout << "0" << endl;
            }
            else cout << (n-a*mini)/lcm+1 << endl;//因为每次都转移lcm 
        }
    }
    return 0;
}