uva 10627

我覺得還蠻難的一道題目 WA折騰了很久

要利用一些技巧，精度問題，

 

初始想法是當快的人跑一次 一定會撞一次

 

除了直接在端點上面撞上，而且有可能會有不同端點的問題

//
//        GGGGGGGGGGGGG        CCCCCCCCCCCCC               AAA
//     GGG::::::::::::G     CCC::::::::::::C              A:::A
//   GG:::::::::::::::G   CC:::::::::::::::C             A:::::A
//  G:::::GGGGGGGG::::G  C:::::CCCCCCCC::::C            A:::::::A
// G:::::G       GGGGGG C:::::C       CCCCCC           A:::::::::A
//G:::::G              C:::::C                        A:::::A:::::A
//G:::::G              C:::::C                       A:::::A A:::::A
//G:::::G    GGGGGGGGGGC:::::C                      A:::::A   A:::::A
//G:::::G    G::::::::GC:::::C                     A:::::A     A:::::A
//G:::::G    GGGGG::::GC:::::C                    A:::::AAAAAAAAA:::::A
//G:::::G        G::::GC:::::C                   A:::::::::::::::::::::A
// G:::::G       G::::G C:::::C       CCCCCC    A:::::AAAAAAAAAAAAA:::::A
//  G:::::GGGGGGGG::::G  C:::::CCCCCCCC::::C   A:::::A             A:::::A
//   GG:::::::::::::::G   CC:::::::::::::::C  A:::::A               A:::::A
//     GGG::::::GGG:::G     CCC::::::::::::C A:::::A                 A:::::A
//        GGGGGG   GGGG        CCCCCCCCCCCCCAAAAAAA                   AAAAAAA
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <climits>
#include <vector>

#include <set>
#include <map>
#include <queue>
#include <cctype>
#include <utility>
#include <ctime>
using namespace std;
#ifdef DEBUG
#define debug(...) printf("!__DEBUG: "),printf(__VA_ARGS__)
#define gettime() end_time=clock();printf("now running time is %.7f\n",(float)(end_time - start_time)/CLOCKS_PER_SEC);
#else
#define debug(...)
#define gettime()
#endif
typedef unsigned int uint;
typedef long long int Int;
#define Set(a,s) memset(a,s,sizeof(a))
#define Write(w) freopen(w,"w",stdout)
#define Read(r) freopen(r,"r",stdin)
#define Pln() printf("\n")
#define I_de(x,n)for(int i=0;i<n;i++)printf("%d ",x[i]);Pln()
#define De(x)printf(#x"%d\n",x)
#define For(i,x)for(int i=0;i<x;i++)
#define CON(x,y) x##y
#define Pmz(dp,nx,ny)for(int hty=0;hty<ny;hty++){for(int htx=0;htx<nx;htx++){\
    printf("%d ",dp[htx][hty]);}Pln();}
#define M 55
#define PII pair<int,int>
#define PB push_back
#define oo INT_MAX
#define Set_oo 0x3f
#define FOR(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
#define eps 1e-10
bool xdy(double x,double y){return x>y+eps;}
bool xddy(double x,double y){return x>y-eps;}
bool xcy(double x,double y){return x<y-eps;}
bool xcdy(double x,double y){return x<y+eps;}
bool xeqy(double x,double y){return fabs(x-y)<eps;}
int min3(int x,int y,int z){
    int tmp=min(x,y);
    return min(tmp,z);
}
int max3(int x,int y,int z){
    int tmp=max(x,y);
    return max(tmp,z);
}
Int l,u,v,t;
Int gcd(Int a,Int b){
    if(b==0)return a;
    return gcd(b,a%b);
}
void solve(){
    Int g=gcd(u,v);
//    debug("%lld %lld\n",u,v);
    if(v==0){
        if(u==0){
            puts("0");
            return ;
        }
        Int ans=t*u/l;
//        debug("%lld\n",ans);
        printf("%lld\n",ans/2+((ans&1)));
        return ;
    }
    long double su=1.0*l/u,sv=1.0*l/v;
    Int ans=(t*u)/l;
    Int ans2=(t*v)/l;
    Int fans=ans;
    long double mt=1.0*l/g;
//    double r=t-ans*su;
//    double r2=t-ans2*sv;
    debug("%lld\n",fans);
    Int tm=t*u%l;
    Int tm2=t*v%l;
    tm%=l;tm2%=l;
        debug("tm>>%lld %lld %lld %lld\n",tm,tm2,ans,ans2);
    if(((ans+ans2)&1)&&(tm!=0||tm2!=0)){
        if(tm>=tm2)fans++;
    }else{
        if(tm>=l-tm2)fans++;
    }
    debug("%lld\n",fans);
    if((u/g-v/g)&1){
        Int tmp=t/mt;
        debug("tmp >>%lld %lld %f \n",tmp,t,mt);
        fans-=tmp/2;
        if(xdy(t,tmp*mt)&&(tmp&1)){
            fans-=1;
        }
    }
//    debug("%lld %.6f\n\n",fans,mt);
    printf("%lld\n",fans);
}
int main() {
//    freopen("out.txt","w",stdout);
    ios_base::sync_with_stdio(0);
    while(~scanf("%lld%lld%lld%lld",&l,&u,&v,&t)&&l){
        if(v>u)swap(u,v);
        solve();

    }
//        gettime();

















}