Ultraviolet A 11768

Ultraviolet A 11768-Lattice Point or Not

Question Link

Given two vertices to form a line segment, these points are in the form of a very bit, and the positive number points fall on this line.

Train of Thought: first form a x + B y = c for the linear table, and Form a, B, and c as integers, then use the extended gcd to find the general solutions of x and y, then we know min (x1, x2) <= x <= max (x1, x2), min (y1, y2) <= y <= max (y1, y2 ), in this way, we can find the t range in the general solution, and the integer that t can take is the integer solution. Then we get the answer.

It is worth noting that when a straight line is in a parallel coordinate system, you must make a special judgment.

Code:

#include <stdio.h>#include <string.h>#include <math.h>#include <algorithm>using namespace std;const long long INF = 0x3f3f3f3f3f3f3f;int t;long long xx1, yy1, xx2, yy2;long long a, b, c;long long read(){double t;scanf("%lf", &t);return (long long)(10 * (t + 0.05));}long long gcd(long long a, long long b) {if (!b) return a;return gcd(b, a % b);}long long exgcd(long long a, long long b, long long &x, long long &y) {if (!b) {x = 1; y = 0; return a;}long long d = exgcd(b, a % b, y, x);y -= a / b * x;return d;}void build() {a = (yy2 - yy1) * 10;b = (xx1 - xx2) * 10;c = (yy2 - yy1) * xx1 + (xx1 - xx2) * yy1;long long t = gcd(gcd(a, b), c);a /= t; b /= t; c /= t;}long long solve() {long long ans = 0;long long x, y;long long d = exgcd(a, b, x, y);long long up = INF, down = -INF;if (xx1 > xx2) swap(xx1, xx2);if (yy1 > yy2) swap(yy1, yy2);if (c % d) return ans;if (b / d > 0) {down = max(down, (long long)ceil((xx1 * d * 1.0 / 10 - x * c * 1.0) / b));up = min(up, (long long)floor((xx2 * d * 1.0 / 10 - x * c * 1.0) / b)); } else if (b / d < 0) { up = min(up, (long long)floor((xx1 * d * 1.0 / 10 - x * c * 1.0) / b));down = max(down, (long long)ceil((xx2 * d * 1.0 / 10 - x * c * 1.0) / b));  }  else if (xx1 % 10) return ans;  if (a / d > 0) {  down = max(down, (long long)ceil((y * c * 1.0 - d * yy2 * 1.0 / 10) / a));  up = min(up, (long long)floor((y * c * 1.0 - d * yy1 * 1.0 / 10) / a));   }   else if (a / d < 0) {   up = min(up, (long long)floor((y * c * 1.0 - d * yy2 * 1.0 / 10) / a));  down = max(down, (long long)ceil((y * c * 1.0 - d * yy1 * 1.0 / 10) / a));   }   else if (yy1 % 10) return ans;   if (down <= up)   ans += up - down + 1;return ans;}int main() {scanf("%d", &t);while (t--) {xx1 = read(); yy1 = read(); xx2 = read(); yy2 = read();build();printf("%lld\n", solve()); }return 0;}