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。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 typedef long long ll; 5 ll ex_gcd(ll a, ll b, ll &x, ll &y) { 6 if(b == 0) { 7 x = 1, y = 0; 8 return a; 9 } 10 ll r = ex_gcd(b, a%b, y, x); 11 y -= (a/b)*x; 12 return r; 13 } 14 int main() { 15 int T; 16 scanf("%d", &T); 17 while(T--) { 18 int n, a, b; 19 ll x=0, y=0; 20 scanf("%d%d%d", &n, &a, &b); 21 ll c = n + 1LL; 22 int t = __gcd(a, b); 23 if(c%t != 0) { 24 printf("0 "); continue; 25 } 26 c /= t, a /= t, b /= t; 27 ex_gcd(a, b, x, y); 28 x *= c, y *= c; 29 ll x1, x2, y1, y2; 30 if(x <= 0) { 31 ll o = (x+1)/b - 1, p = y/a; 32 x1 = x - o * b, y1 = y + o * a; 33 x2 = x + p * b, y2 = y - p * a; 34 } else { 35 ll o = x/b, p = (y+1)/a - 1; 36 x1 = x - o * b, y1 = y + o * a; 37 x2 = x + p * b, y2 = y - p * a; 38 } 39 ll ans = abs(x1-x2)/b + 1; 40 ans -= (x1<=0||y1<=0) + (x2<=0||y2<=0); 41 printf("%lld ", max(0LL, ans)); 42 } 43 return 0; 44 }