标签：game 技术 ica long stream a* int follow i++
There is an interesting and simple one person game. Suppose there is a number axis under your feet. You are at point A at first and your aim is point B. There are 6 kinds of operations you can perform in one step. That is to go left or right by a,b and c, here c always equals to a+b.
You must arrive B as soon as possible. Please calculate the minimum number of steps.
There are multiple test cases. The first line of input is an integer T(0 < T ≤ 1000) indicates the number of test cases. Then T test cases follow. Each test case is represented by a line containing four integers 4 integers A, B, a and b, separated by spaces. (-231 ≤ A, B < 231, 0 < a, b < 231)
For each test case, output the minimum number of steps. If it‘s impossible to reach point B, output "-1" instead.
2 0 1 1 2 0 1 2 4
1 -1
Author: CAO, Peng
Contest: The 12th Zhejiang University Programming Contest
1 #include <iostream> 2 #include <cstring> 3 #include <algorithm> 4 using namespace std; 5 6 typedef long long ll; 7 ll t; 8 ll d1,d2,d3,d4; 9 ll X,Y,a,b,c; 10 ll GDC; 11 12 ll gdc(ll x,ll y){ 13 return y==0?x:gdc(y,x%y); 14 } 15 16 void extend_gdc(ll a,ll b,ll &X,ll &Y){ 17 if(b==0){ 18 X=1; 19 Y=0; 20 return; 21 } 22 extend_gdc(b,a%b,X,Y); 23 ll temp=X; 24 X=Y; 25 Y=temp-a/b*Y; 26 } 27 28 int main(){ 29 ios::sync_with_stdio(false); 30 while(cin>>t){ 31 while(t--){ 32 cin>>d1>>d2>>d3>>d4; 33 a=d3,b=d4,c=d2-d1; 34 GDC=gdc(a,b); 35 if(c%GDC!=0){ 36 cout << "-1" << endl; 37 continue; 38 } 39 extend_gdc(a,b,X,Y); 40 X=X*c/GDC; 41 Y=Y*c/GDC; 42 a=a/GDC; 43 b=b/GDC; 44 ll mid=(Y-X)/(a+b); 45 ll res=0x3f3f3f3f3f3f3f3f; 46 for(int i=mid-1;i<=mid+1;i++){ 47 ll ans=0; 48 ll ee=X+b*i; 49 ll rr=Y-a*i; 50 if(ee*rr>=0) ans+=max(abs(ee),abs(rr)); //同号取最大值 51 else ans+=abs(ee)+abs(rr); //异号模相加 52 // cout << ans << endl; 53 res=min(res,ans); 54 } 55 cout << res << endl; 56 } 57 } 58 return 0; 59 }
标签：game 技术 ica long stream a* int follow i++
原文地址：https://www.cnblogs.com/qq-1585047819/p/11336228.html
