UVA 11768 Lattice Point or Not

扩展欧几里得，给两个点，就可以求出直线方程为 (yy-y)*x0 + (x-xx)*y0 = x*yy - y*xx，求的是在线段上的整点个数。所以就是(yy-y)*10*x0 + (x-xx)*10*y0 = x*yy - y*xx满足条件的解的个数。用exgcd搞之后求出一个解，再求出在线段上第一个整点的位置，然后再求有多少个在线段上的点。

exgcd有点忘了，还有就是特殊情况的判断（比如平行坐标轴），另外就是不能交换输入点，输入

1.0 0.3 0.3 10.0交换后就是0.3 0.3 1.0 10.0就变成有整点了，下次一定要注意，多想想

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<string>
#include<algorithm>

using namespace std;
#define inf 0x3f3f3f3f
#define eps 1e-8
#define LL long long
#define ull unsigned long long
#define mnx 1005

void exgcd( LL a, LL b, LL &d, LL &x, LL &y ){
    if( !b ) {
        d = a, x = 1, y = 0;        // d恰好是最大公约数
    }
    else {
        exgcd( b, a%b, d, y, x );
        y -= x*(a/b);
    }
}
double ax, ay, bx, by;
void solve(){
    if( ax > bx ) swap( ax, bx );
    if( ay > by ) swap( ay, by );
    LL x = ax * 10, y = ay * 10, xx = bx * 10, yy = by * 10;
    if( x == xx || y == yy ){
        if( x == xx ) swap( ax, ay ), swap( bx, by ), swap( x, y ), swap( xx, yy );
        if( yy % 10 ){
            puts( "0" ); return ;
        }
        x = ceil( ax ) * 10, xx = floor( bx ) * 10;
        cout << x << " " << xx << endl;
        if( xx < x ){
            puts( "0" ); return ;
        }
        printf( "%lld
", (xx-x)/10 + 1LL ); return ;
    }
    LL a = ( yy - y ) * 10, b = ( x - xx ) * 10, c = x * yy - y * xx, d, x0, y0;
    //cout << a << " "<< b <<endl;
    exgcd( a, b, d, x0, y0 );
    //cout <<c <<" " << d << endl;
    if( c % d ){
        puts( "0" ); return ;
    }
    x = ceil( ax ), xx = floor( bx );
    y = ceil( ay ), yy = floor( by );
    //cout << 1 << endl;
    if( x > xx || y > yy ){
        puts( "0" ); return ;
    }
    x0 = x0 * ( c / d );
    //cout << x0 << endl;
    b /= d;
    if( b < 0 ) b = -b;
    x0 = x0 - ( x0 - x ) / b * b;
    x0 -= b;
    while( x0 < x ) x0 += b;
    LL ans = ( xx - x0 ) / b;
    while( x0 + ans * b <= xx ) ans++;
    printf( "%lld
", ans );
}
int main(){
    int cas;
    scanf( "%d", &cas );
    while( cas-- ){
        scanf( "%lf%lf%lf%lf", &ax, &ay, &bx, &by );
        solve();
    }
    return 0;
}