51nod

题目链接：1352 集合计数

思路：本题相当于求 n+1 = Ax + By。求（x, y）。（x>0, y>0)

用扩展GCD求出一组解之后，通过这组解找到x>0的最小的解和y>0的最小的解，就可以求出一共有多少组解了。

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
ll ex_gcd(ll a, ll b, ll &x, ll &y) {
    if(b == 0) {
        x = 1, y = 0;
        return a;
    }
    ll r = ex_gcd(b, a%b, y, x);
    y -= (a/b)*x;
    return r;
}
int main() {
    int T;
    scanf("%d", &T);
    while(T--) {
        int n, a, b;
        ll x=0, y=0;
        scanf("%d%d%d", &n, &a, &b);
        ll c = n + 1LL;
        int t = __gcd(a, b);
        if(c%t != 0) {
            printf("0
"); continue;
        }
        c /= t, a /= t, b /= t;
        ex_gcd(a, b, x, y);
        x *= c, y *= c;
        ll x1, x2, y1, y2;
        if(x <= 0) {
            ll o = (x+1)/b - 1, p = y/a;
            x1 = x - o * b, y1 = y + o * a;
            x2 = x + p * b, y2 = y - p * a;
        } else {
            ll o = x/b, p = (y+1)/a - 1;
            x1 = x - o * b, y1 = y + o * a;
            x2 = x + p * b, y2 = y - p * a;
        }
        ll ans = abs(x1-x2)/b + 1;
        ans -= (x1<=0||y1<=0) + (x2<=0||y2<=0);
        printf("%lld
", max(0LL, ans));
    }
    return 0;
}