Good questions, interesting questions, and troublesome questions

Sgu 106:

First, let's solve a linear equation AX + by =-C, which can be done by extending Euclidean. However, it requires us to output the number of solutions that meet X1 <= x <= x2, Y1 <= Y <= Y2, which is not that simple...

At first I wanted to record the boundary values of X and Y, But I thought for a long time and divided them into n cases... because it was too cumbersome, I simply gave up the score and handed in a wa.

After reading someone's problem-solving report, we found that we can find the number of K that satisfies X1 <= x0 + K * R1 <= x2, Y1 <= y0-k * R2 <= Y2. R1 = LCM/a, R2 = LCM/B, LCM = a * B * gcd (A, B ).

Then, solve the equation and calculate the intersection length of the interval.

(The Ceil and floor functions need to be # include <cmath>, while the cmath contains a function named Y1, which is really painful. But in fact, you only need to get Y1 from the global system to the inside of the main function)

1 # include <iostream> 2 # include <cstdio> 3 # include <cmath> 4 # define debug (X) cout <# x <"=" <x <Endl 5 using namespace STD; 6 typedef long ll; 7 8 LL gcd (ll a, LL B) 9 {10 LL temp; 11 if (a <0) A =-A; If (B <0) B =-B; 12 while (B! = 0) {temp = B; B = A % B; A = temp;} 13 return a; 14} 15 // input, B must be mutually qualitative (because at the end of a * 1 + B * 0 = 1, ensure that A is 1) 16 void _ gcd (ll a, LL B, ll & X, ll & Y) 17 {18 if (B = 0) {x = 1, y = 0; return;} 19 _ gcd (B, A % B, X, y); 20 ll temp = y; 21 y = x-A/B * Y; X = temp; 22} 23 int main () 24 {25 ll A, B, c, x1, x2, Y1, Y2; 26 ll X, Y; 27 ll tx1, tx2, ty1, ty2; 28 CIN> A> B> C> x1> X2> Y1> Y2; 29 ll r = gcd (a, B), R1, R2; 30 if (a = 0 & B = 0) 31 {32 If (C = 0) cout <(x2-x1 + 1) * (y2-y1 + 1) <Endl; 33 else cout <0 <Endl; 34} 35 else if (a = 0 | B = 0) 36 {37 if (a = 0) Swap (a, B), swap (x1, Y1), swap (X2, Y2); 38 If (C %! = 0) cout <0 <Endl; 39 else if (-C/A> = x1 &-C/A <= x2) cout <(y2-y1 + 1) <Endl; 40 else cout <0 <Endl; 41} 42 else if (C % R! = 0) printf ("0 \ n"); 43 else44 {45 _ gcd (A/R, B/R, x, y ); 46 X * =-C/R; y * =-C/R; 47 r = a * B/R; R1 = r/a; r2 = r/B; 48 tx1 = x1-x; tx2 = x2-x; If (R1 <0) Swap (tx1, tx2), tx1 =-tx1, tx2 =-tx2, R1 =-R1; 49 ty1 =-y2 + Y; ty2 =-Y1 + Y; If (R2 <0) Swap (ty1, ty2), ty1 =-ty1, ty2 =-ty2, r2 =-R2; 50 tx1 = Ceil (double (tx1)/R1); tx2 = floor (double (tx2)/R1); 51 ty1 = Ceil (double (ty1) /R2); ty2 = floor (double (ty2)/R2); 52 tx1 = max (tx1, ty1); tx2 = min (tx2, ty2 ); 53 If (tx2> = tx1) cout <tx2-tx1 + 1 <Endl; 54 else cout <0 <Endl; 55} 56 return 0; 57}

The equation

 

Good questions, interesting questions, and troublesome questions