CodeForces Gym101161D Find C

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Source:2016-2017 ACM-ICPC Asia-Bangkok Regional Contest

Problem:二维平面，给两个整数点A,B，找k个整数点C使得ABC是个非退化三角形，并且三角形的内部和边上都没有整数点。

Idea:题目转化为找到离这条直线最近的有整数点的平行线，显然ax+by=gcd(a ,b)这条直线是离ax+by=0这条直线最近的直线，那么找k组解并平移一下即可。

Code:

#include<bits/stdc++.h>using namespace std;#define fi first#define se second#define pb push_back#define lson o<<1#define rson o<<1|1#define CLR(A, X) memset(A, X, sizeof(A))#define bitcount(X) __builtin_popcountll(X)typedef long long LL;typedef pair<int, int> PII;const double eps = 1e-10;const double PI = acos(-1.0);const LL MOD = 1e9+7;const auto INF = 0x3f3f3f3f;int dcmp(double x) { if(fabs(x) < eps) return 0; return x<0?-1:1; }const int MAXN = 1e5+5;void exgcd(LL a, LL b, LL &d, LL &x, LL &y) {    if(!b) { d = a; x = 1; y = 0; }    else { exgcd(b, a%b, d, y, x); y -= x*(a/b); }}int main() {    int X;    scanf("%d", &X);    while(X--) {        LL ax, ay, bx, by, x, y, g;        int k;        scanf("%lld%lld%lld%lld%d", &ax, &ay, &bx, &by, &k);        LL a = by-ay, b = ax-bx;        exgcd(a, b, g, x, y);        a /= g, b /= g;        while(k--) {            x += b;            y -= a;            printf("%lld %lld\n", ax+x, ay+y);        }    }    return 0;}

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