SGU，人品測試器

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做sgu上的106時，總是WA on TEST#11, 後來參考某牛的標程，把y0 *= -c 改成了  y0 = (-c - (long long)a * x0) / b ，兩者理論上效果一樣的，而且前者理論上不可能溢出，但事實是改了後就AC了。。。
附上代碼：

/* * $File: 106.cpp * $Date: Sat Aug 15 12:42:02 2009 */ #include <iostream> #include <cmath> using namespace std; namespace Solve { long long work(int a, int b, int c, int x1, int x2, int y1, int y2); int gcd(int a, int b); void solve_eqn(int a, int b, int &x, int &y); // a * x + b * y = 1 void calc_range(int a, int b, int x1, int x2, int &min, int &max); // x1 <= a + b * x <= x2 void solve(istream &instr, ostream &outstr); } void Solve::solve(istream &instr, ostream &outstr) { int a, b, c, x1, x2, y1, y2; instr >> a >> b >> c >> x1 >> x2 >> y1 >> y2; outstr << work(a, b, c, x1, x2, y1, y2) << endl; } long long Solve::work(int a, int b, int c, int x1, int x2, int y1, int y2) { if (a == 0 || b == 0) { if (a + b == 0) if (c == 0) return (long long)(x2 - x1 + 1) * (y2 - y1 + 1); else return 0; if (a == 0) return c % b == 0 && c / b >= y1 && c / b <= y2; return c % a == 0 && c / a >= x1 && c / a <= x2; } int g = gcd(a, b); if (c % g) return 0; a /= g; b /= g; c /= g; int x0, y0; solve_eqn(a, b, x0, y0); x0 *= -c; y0 = (-c - (long long)a * x0) / b; // not y0 *= -c; // !!!!!!!!!!!!!!!!!!!!!!!!! RP !!!!!!!!!!!!!!!!!!!!!!!! int k1_min, k1_max; calc_range(x0, b, x1, x2, k1_min, k1_max); int k2_min, k2_max; calc_range(y0, -a, y1, y2, k2_min, k2_max); if (k2_min > k1_min) k1_min = k2_min; if (k2_max < k1_max) k1_max = k2_max; if (k1_min > k1_max) return 0; return k1_max - k1_min + 1; } void Solve::calc_range(int a, int b, int x1, int x2, int &min, int &max) { double m, n; m = x1 - a; n = x2 - a; // m <= b * x <= n double min1, max1; if (b < 0) { min1 = n / b; max1 = m / b; } else { min1 = m / b; max1 = n / b; } min = (int)ceil(min1); max = (int)floor(max1); } void Solve::solve_eqn(int a, int b, int &x, int &y) { bool minus_a = (a < 0), minus_b = (b < 0); if (minus_a) a = -a; if (minus_b) b = -b; x = 1; int g = a, v = 0, w = b; while (w) { int t = g % w, q = g / w; int s = x - q * v; x = v; v = s; g = w; w = t; } y = (g - a * x) / b; if (minus_a) x = -x; if (minus_b) y = -y; } int Solve::gcd(int a, int b) { if (a < 0) a = -a; if (b < 0) b = -b; while (b) { int tmp = a % b; a = b; b = tmp; } return a; } int main() { Solve::solve(cin, cout); }