欧几里德算法

上次做2017 apac test 最后一轮的时候，第三题需要使用扩展欧几里德算法来求解方程的解的个数，刚好学习一下这个算法。上次学了一下，忘记记录下来，这次在做hr的题目的时候，有需要这个算法，还是不熟练，这里还是把上次的代码，先贴出来吧，需要的进行研究！

#include<bits/stdc++.h>
#define pb push_back
#define FOR(i, n) for (int i = 0; i < (int)n; ++i)
#define dbg(x) cout << #x << " at line " << __LINE__ << " is: " << x << endl
typedef long long ll;
using namespace std;
typedef pair<int, int> pii;
const int maxn = 1e3 + 10;
int a, b;
int gcd(int x, int y) {
    if(y == 0) return x;
    return gcd(y, x % y);
}
int gcd2(int x, int y) {
    while(y != 0) {
        int t = x % y;
        x = y;
        y = t;
    }
    return x;
}
int exgcd(int a, int b, int &x, int &y) {
    if(b == 0) {
        x = 1;
        y = 0;
        return a;
    }
    int r = exgcd(b, a % b, x, y);
    int t = x;
    x = y;
    y = t - a / b * y;
    return r;
}
int work(int a, int b, int c) {
    int x, y;
    ll t = exgcd(a, b, x, y);
    //cout << t << " " << x << " " << y << endl;
    if(c % t != 0) return 0;
    ll f = c / t;
    //double z1 = 1.0 * -x * f * t / b;
    //double z2 = 1.0 * y * f * t / a;
    //int x1 = z1, x2 = z2;
    ll t1 = floor(1.0 * -x * f * t / b);
    ll t2 = floor(1.0 * y * f * t / a);
    //cerr << z1 << " " << x1 << " " << t1 << " " << z2 << " " << x2 << " " << t2 << endl;
    if(1ll * y * f * t % a == 0) t2--;
    return t2 - t1 < 0 ? 0 : t2 - t1;
}
/*
cin >> a >> b;
    cout << __gcd(a, b) << endl;
    cout << gcd(a, b) << " " << gcd2(a, b) << endl;
    int x, y;
    int t = exgcd(a, b, x, y);
    cout << t << " " << x << " " << y << endl;
*/
void extgcd(ll a,ll b,ll& d,ll& x,ll& y){
    if(!b){
        d=a;x=1;y=0;
    }else{
        extgcd(b,a%b,d,y,x);
        y-=x*(a/b);
    }
}

int work1(int a,int b,int n){
    if(n%__gcd(a,b)!=0)return 0;
    ll g,p0,q0;
    extgcd(a,b,g,p0,q0);
    p0*=n/g;q0*=n/g;
    int less=floor((-p0)*g*1.0/b),great=floor(q0*g*1.0/a);
    if(q0*g%a==0)--great;
    return great-less<0?0:great-less;
}
void solve1() {
    cin >> a >> b;
    ll res = 0;
    //if(a % b == 0) res++;
    for (int i = b; i <= a; i += b) {
        int now = a;
        if(now % i == 0) res++;
        res += work(i, i + 1, now);
        res += work(i, i + 2, now);
        for (int j = 1; ; j++) {
            int t = now - i * j;
            if(t <= 0) break;
            res += work(i + 1, i + 2, t);
        }
        //cout << i << " " << res << endl;
    }
    cout << res << endl;
}
ll work2(int a, int b, int c) {
    int x, y;
    ll t = exgcd(a, b, x, y);
    //cout << t << " " << x << " " << y << endl;
    if(c % t != 0) return 0;
    ll f = c / t;
    //double z1 = 1.0 * -x * f * t / b;
    //double z2 = 1.0 * y * f * t / a;
    //int x1 = z1, x2 = z2;
    ll t1 = floor(1.0 * -x * f * t / b);
    ll t2 = floor(1.0 * y * f * t / a);
    //cerr << z1 << " " << x1 << " " << t1 << " " << z2 << " " << x2 << " " << t2 << endl;
    if(1ll * -x * f * t % b == 0) t1--;
    return t2 - t1 < 0 ? 0 : t2 - t1;
}
void solve() {
    cin >> a >> b;
    ll res = 0;
    //if(a % b == 0) res++;
    for (int i = b; i <= a; i += b) {
        //int now = a - b;
        for (int j = 1; ; j++) {
            int now = a - i * j;
            if(now < 0) break;
            res += work2(i + 1, i + 2, now);
        }

        //cout << i << " " << res << endl;
    }
    cout << res << endl;
}
int main() {
    freopen("test.in", "r", stdin);
    freopen("test.out", "w", stdout);
    //freopen("test.err", "w", stderr);
    //cout << __builtin_popcount(11) << endl;
    //return 1;
    //cout << work(2, 3, 5) << endl;
    //return 0;
    int _;
    cin >> _;
    for (int i = 1; i <= _; i++) {
        cout << "Case #" << i << ": ";
        solve();
    }
    return 0;
}