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UVA 11768 Lattice Point or Not

2024-07-28 07:50:46   215人阅读

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 0x3f3f3f3f11 #define eps 1e-812 #define LL long long13 #define ull unsigned long long14 #define mnx 100515 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 }
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UVA 11768 Lattice Point or Not


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