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UVA 11768 - Lattice Point or Not(数论)
UVA 11768 - Lattice Point or Not
题目链接
题意:给定两个点,构成一条线段,这些点都是十分位形式的,求落在这个直线上的正数点。
思路:先把直线表达成a x + b y = c的形式,a,b, c都化为整数表示,然后利用扩展gcd求出x和y的通解,然后已知min(x1, x2) <= x <= max(x1, x2), min(y1, y2) <= y <= max(y1, y2),这样一来就可以求出通解中t的范围,t能取的整数就是整数解,就得到答案。
值得注意的是,直线为平行坐标系的情况,要特殊判断一下
代码:
#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; }
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