zoj 4056

• At 0 second, the LED light is initially off. After BaoBao presses the button 2 times, the LED light turns on and the value of the counter changes to 1. The value of the timer is also set to 2.5 seconds. After DreamGrid presses the button 1 time, the value of the counter changes to 2.

• At 2.5 seconds, the timer counts down to 0 and the LED light is off.

• At 5 seconds, after DreamGrid presses the button 1 time, the LED light is on, and the value of the timer is set to 2.5 seconds.

• At 7.5 seconds, the timer counts down to 0 and the LED light is off.

• At 8 seconds, after BaoBao presses the button 2 times, the LED light is on, the value of the counter changes to 3, and the value of the timer is set to 2.5 seconds.

• At 10 seconds, after DreamGrid presses the button 1 time, the value of the counter changes to 4, and the value of the timer is changed from 0.5 seconds to 2.5 seconds.

• At 12.5 seconds, the timer counts down to 0 and the LED light is off.

• At 15 seconds, after DreamGrid presses the button 1 time, the LED light is on, and the value of the timer is set to 2.5 seconds.

• At 16 seconds, after BaoBao presses the button 2 times, the value of the counter changes to 6, and the value of the timer is changed from 1.5 seconds to 2.5 seconds.

• At 18 seconds, the game ends.

只要遇到a,c的倍数就会按b,d次按钮，若同是a,c的倍数，先a,后c
若按之前为暗，按一次变亮，若按之前为亮，按一次为计数器+1
每次按一下，都会让计时器重新设置为v+0.5（开始倒计时，时间到0，就会变暗）
问[0,t]的时间内，计数器最后为多少。(0为任意数的倍数）

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <string>
 #include <cmath>
 using  namespace std;
 #define ll long long
 int T;
 ll a,b,c,d,v,t;
 ll gcd(ll a,ll b)
 {
     return b==?a:gcd(b,a%b);
 }
 ll lcm(ll a,ll b)
 {
     return a*b/gcd(a,b);
 }
 ll x,y,z,cnt,ans,last;
 /*
 明显一个最小公倍数为一个周期
 0 :单独算
 再计算除一个周期的结果*周期数
 再加上不满一个周期的结果
 */
 int  main()
 {
     scanf("%d",&T);
     while(T--){
     scanf("%lld%lld%lld%lld%lld%lld",&a,&b,&c,&d,&v,&t);
     ans=b+d-;
     x=,y=,last=,cnt=;//ans 不用再赋初值了
     /*

     */
     z=lcm(a,c);//(a,c) 不是(x,y)
     while(x<z||y<z){
         if(x+a<=y+c){
             x+=a;
             if(x<=last+v) cnt++;
             cnt+=b-;
             last=x;
         }
         else{
             y+=c;
             if(y<=last+v) cnt++;
             cnt+=d-;
             last=y;
         }

     }
     ans+=cnt*(t/z);//cnt为一个周期的结果
     t%=z;
     x=,y=,last=;
     while(x+a<=t||y+c<=t){//保证在t的范围内
         if(x+a<=y+c){
             x+=a;
             if(x<=last+v) ans++;
             ans+=b-;
             last=x;
         }
         else{
             y+=c;
             if(y<=last+v) ans++;
             ans+=d-;
             last=y;
         }
     }
     printf("%lld\n",ans);
     }
     return  ;
 }