青蛙的约会

青蛙的约会

设跳k次，能遇到一起，说明到了同一个点，(x + k * m) = (y + k * n)    (mod l), k *(n - m) = (x - y) (mod l)

k * (n - m) + t * l = x - y; 令a = n - m; x = k; y = t; b = l; c = x - y;

ax + by = c 注意a,b 必须是非负整数

ax + by = gcd(a,b) = d;得到特解x0 

如果ax + by = c 有解，说明 d | c; x = x0 * c / d;

x最小的非负整数 x = (x0 * c / d % b / d + b / d) % b / d

#include <bits/stdc++.h>
using namespace std;
#define int long long
int x,y,n,m,l,x0,yy,d;
int exgcd(int a,int b,int &x,int &y){
    if(!b){
        x = 1,y = 0;
        return a;
    }
    d = exgcd(b,a % b,y,x);
    y -= x * (a / b);
    return d;
}
signed main(){
    //freopen("in","r",stdin);
    ios::sync_with_stdio(0);
    cin >> x >> y >> m >> n >> l;
    int a = n - m, b = l, c = x - y;
    if(a < 0) a = -a, c = -c;
    exgcd(a,b,x0,yy);
    if(c % d != 0) cout << "Impossible";
    else cout << (x0 * (c / d) % (b / d) + b / d) % (b / d);
    return 0;
}

注：这里面返回d是最大公约数,x0是一个特殊解