Hdu4596 yet another end of the world extends Euclidean Properties

I should have eaten this question. At first, the equation was wrong and I didn't want to think about the problem of connecting columns. I am really confused. If I can't do it, I should repeat it, fuck,

I gave N groups of X, Y, and Z and asked if there are two or more IDs. Yes, the value of ID % x is between the range [y, z, if yes, the output is cannot take off.

Otherwise you know

Based on the meaning of the question:

A * X1 + Y1 <= ID <= A * X1 + z1

B * X2 + y2 <= ID <= B * X2 + Z2,

If there is a solution, there must be duplicates between the two intervals, so continue to push:

A * X1 <= B * X2 + Z2 and B * X2 + y2 <= A * X1 + z1

Simplified to get

A * x1-B * X2 <= Z2-Y1 1 subaccount

B * x2-A * X1 <= Z1-Y2, expression 2

In order to unify the number 2

A * x1-B * X2 <= Y2-z1

Then we can see that we should think of extending Euclidean. If there is a non-negative and not all 0 integers A, B, there must be an integer pair X, Y to make gcd (A, B) = A * x + B * y,

To apply this question, gcd (x1, x2) = A * X1 + (-B * x2) is required, the sub-value of this formula is Z2-Y1 at the same time, so it is actually converted

The relationship between gcd (x1, x2) and Z2-Y1 is an integer multiple, meaning (z2-y1) % gcd (x1, x2) = 0, note that it is a modulo, I saw a question with a division number. When I tried to see other people's practices, I got a question and thought I was wrong, the problem is that someone else directly copied the parsing ~~~~~

At the same time, for Formula 2, we can also push to (Y2-Z1) %) gcd (x1, x2) = 0. either of the two can be satisfied, and the range coverage does not need to be strictly included, it's hard to tell the language clearly. I leave it to my mind.

N groups are given in the question. The range of N is 1000. You can simply scan two layers of violence to determine whether it is OK,

My judgment is as follows:

Int CAL (int A, int B, int c) {If (C % A = 0 | B % A = 0) return 1; // return 0 in case of equality of inequality ;}

The question judgment part of others is:

bool isok(int t,int le,int ri)  {      int i,j;      if(le%t==0||ri%t==0) return true ;      if(le<0&&ri>=0) return true ;      if(ri/t-le/t>0) return true ;      return false ;  }  

The judgment of all the online questions I have seen so far is exactly the same. All of them are as shown above, but I have missed the two if statements below, is the data on the subject too watery? The following two judgments I have not understood yet. Is there anything missing from my weak dish? If you pass by, ask me for advice,

# Include <iostream> # include <cstdio> # include <list> # include <algorithm> # include <cstring> # include <string> # include <queue> # include <stack> # include <map> # include <vector> # include <cmath> # include <memory. h ># include <set> # define ll long # define EPS 1e-8 # define INF 0 xfffffffconst ll INF = 1ll <61; using namespace STD; // vector <pair <int, int> G; // typedef pair <int, int> P; // vector <pair <int, int> >:: iterator ITER;/ /// Map <LL, int> MP; // Map <LL, int>: iterator P; const int n = 1000 + 5; int X [N], Y [N], Z [N]; void Init () {memset (x, 0, sizeof (x); memset (Y, 0, sizeof (y )); memset (z, 0, sizeof (z);} int gcd (int A, int B) {return B = 0? A: gcd (B, A % B);} int CAL (int A, int B, int C) {If (C % A = 0 | B % A = 0) return 1; // return 0 in case of equality of inequality;} int main () {int N; while (scanf ("% d", & n) = 1) {Init (); For (INT I = 0; I <n; I ++) scanf ("% d", & X [I], & Y [I], & Z [I]); bool flag = false; for (INT I = 0; I <n; I ++) {for (Int J = I + 1; j <n; j ++) {int TMP = gcd (X [I], X [J]); If (CAL (TMP, Z [J]-y [I], Y [J]-Z [I]) {flag = true ;}} if (FLAG) puts ("cannot take off "); else puts ("can take off");} return 0 ;}