标签：style class blog c code java
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 integersA, 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
#include<iostream> #include<stdio.h> #include<cstring> #include<cstdlib> #include<math.h> using namespace std; typedef long long LL; LL Ex_GCD(LL a,LL b,LL &x,LL &y) { if(b==0) { x=1; y=0; return a; } LL g=Ex_GCD(b,a%b,x,y); LL hxl=x-(a/b)*y; x=y; y=hxl; return g; } LL get(LL a,LL b) { if(a*b<0) { return abs(a)+abs(b); } else return abs(a)>abs(b)? abs(a):abs(b); } int main() { int T; LL A,B,a,b,g,x,y,ans,cur,c; scanf("%d",&T); while(T--) { scanf("%lld%lld%lld%lld",&A,&B,&a,&b); g=Ex_GCD(a,b,x,y); c=A-B; if(c<0) c=-c; if(c%g!=0) { printf("-1\n"); continue; } a=a/g; b=b/g; x=x*(c/g); y=y*(c/g); double t = (y-x)*1.0/(a+b); LL K = (LL)t; cur = get(x+b*K,y-a*K); ans=cur; K++; cur=get(x+b*K,y-a*K); ans=min(ans,cur); K++; cur=get(x+b*K,y-a*K); ans=min(ans,cur); K=K-3; cur=get(x+b*K,y-a*K); ans=min(ans,cur); K=K-1; cur=get(x+b*K,y-a*K); ans=min(ans,cur); printf("%lld\n",ans); } return 0; }
ZOJ 3593 One Person Game,布布扣,bubuko.com
标签：style class blog c code java
原文地址：http://www.cnblogs.com/tom987690183/p/3736363.html
