不定方程与扩展欧几里得

时间不多了，先把代码放上来。

#include<iostream>
#include<cstring>
#include<cstdio>

using namespace std;

typedef long long ll;

ll gcd(ll a, ll b)
{
    return (b == 0)?a:(gcd(b, a%b));
 } 

ll exgcd(ll a, ll b, ll &x, ll &y)
{
    if(b == 0){
        x = 1; y = 0;
        return a;
    }
    ll t = exgcd(b, a%b, y, x);
    y -= (a/b)*x;
    
    return t;
 } 

int main()
{
    ll x, y;
    ll a, b, m;
    cin>>a>>b>>m;
    // 有解的充要条件 
    if(m%gcd(a, b) == 0) cout<<"Yes
"; 
    ll s = exgcd(a, b, x, y);
    x = x*m/s;
    y = y*m/s;
    for(int t = -100;t<=300;t++){
        printf("%lld*(%lld) + %lld*(%lld) = %lld
", a, x, b, y, m);
        /*
        x -= b/s;
        y += a/s;
        */
        // 通解, 可往左右走 
        x -= b/s;
        y += a/s;
    }
    
    return 0;
}