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

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cstdio>
 5 #include<cmath>
 6 #include<string>
 7 #include<algorithm>
 8
 9 using namespace std;
10 #define inf 0x3f3f3f3f
11 #define eps 1e-8
12 #define LL long long
13 #define ull unsigned long long
14 #define mnx 1005
15
16 void exgcd( LL a, LL b, LL &d, LL &x, LL &y ){
17     if( !b ) {
18         d = a, x = 1, y = 0;        // d恰好是最大公约数
19     }
20     else {
21         exgcd( b, a%b, d, y, x );
22         y -= x*(a/b);
23     }
24 }
25 double ax, ay, bx, by;
26 void solve(){
27     if( ax > bx ) swap( ax, bx );
28     if( ay > by ) swap( ay, by );
29     LL x = ax * 10, y = ay * 10, xx = bx * 10, yy = by * 10;
30     if( x == xx || y == yy ){
31         if( x == xx ) swap( ax, ay ), swap( bx, by ), swap( x, y ), swap( xx, yy );
32         if( yy % 10 ){
33             puts( "0" ); return ;
34         }
35         x = ceil( ax ) * 10, xx = floor( bx ) * 10;
36         cout << x << " " << xx << endl;
37         if( xx < x ){
38             puts( "0" ); return ;
39         }
40         printf( "%lld\n", (xx-x)/10 + 1LL ); return ;
41     }
42     LL a = ( yy - y ) * 10, b = ( x - xx ) * 10, c = x * yy - y * xx, d, x0, y0;
43     //cout << a << " "<< b <<endl;
44     exgcd( a, b, d, x0, y0 );
45     //cout <<c <<" " << d << endl;
46     if( c % d ){
47         puts( "0" ); return ;
48     }
49     x = ceil( ax ), xx = floor( bx );
50     y = ceil( ay ), yy = floor( by );
51     //cout << 1 << endl;
52     if( x > xx || y > yy ){
53         puts( "0" ); return ;
54     }
55     x0 = x0 * ( c / d );
56     //cout << x0 << endl;
57     b /= d;
58     if( b < 0 ) b = -b;
59     x0 = x0 - ( x0 - x ) / b * b;
60     x0 -= b;
61     while( x0 < x ) x0 += b;
62     LL ans = ( xx - x0 ) / b;
63     while( x0 + ans * b <= xx ) ans++;
64     printf( "%lld\n", ans );
65 }
66 int main(){
67     int cas;
68     scanf( "%d", &cas );
69     while( cas-- ){
70         scanf( "%lf%lf%lf%lf", &ax, &ay, &bx, &by );
71         solve();
72     }
73     return 0;
74 }