hackerrank Lucky Numbers（扩展gcd/规律）

题目：https://www.hackerrank.com/contests/hourrank-16/challenges/leonardo-and-lucky-numbers/problem
题意：判断N是否是4和7的和
思路：
* 1. 7*x+4*y=N 扩展gcd
* 2.两边都除以4，下图为官方题解
*
7*Y <= N
7*0%4 = 0 N>=0
7*1%4 = 1 N>=7
7*2%4 = 2 N>=14
7*3%4 = 3 N>=21
代码：

#include<iostream>using namespace std;typedef long long ll;ll exGcd(ll a,ll b,ll &x,ll &y){    if(b == 0)    {        x = 1; y = 0;        return a;    }    ll r = exGcd(b,a%b,x,y);    ll t = x;    x = y;    y = t - a/b*y;    return r;}int main(){    int q;    ll n;    cin >> q;    while(q--)    {        cin >> n;//7*x+4*y=n        ll x = 0,y = 0;        ll d = exGcd(7,4,x,y);        if(n % d != 0)            cout << "No" << "\n";        else        {            x = (x*n%4 + 4) % 4;            y = (n - 7*x) / 4;            if(y >= 0)                cout << "Yes" << "\n";            else                cout << "No" << "\n";        }    }    return 0;}