The 2018 ACM-ICPC Asia Qingdao Regional Contest, Online J Press the Button

BaoBao and DreamGrid are playing a game using a strange button. This button is attached to an LED light (the light is initially off), a counter and a timer and functions as follows:

When the button is pressed, the timer is set to
When the button is pressed with the LED light off, the LED light will be lit up;
When the button is pressed with the LED light on, the value of the counter will be increased by 1;
When the timer counts down to 0, the LED light goes out (that is to say, the light is off).

During the game, BaoBao and DreamGrid will press the button periodically. If the current real time (that is to say, the time elapsed after the game starts, NOT the value of the timer) in seconds is an integer and is a multiple of a given integer

Note that

0 is a multiple of every integer;
Both BaoBao and DreamGrid are good at pressing the button, so it takes no time for them to finish pressing;
If BaoBao and DreamGrid are scheduled to press the button at the same second, DreamGrid will begin pressing the button

The game starts at 0 second and ends after

Input

There are multiple test cases. The first line of the input contains an integer

The first and only line contains six integers

Output

For each test case output one line containing one integer, indicating the value of the counter when the game ends.

Sample Input

2
8 2 5 1 2 18
10 2 5 1 2 10

Sample Output

6
4

Hint

We now explain the first sample test case.

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.

求出a和c的最小公倍数，所以0到最小公倍数之间即为一循环节，暴力跑出循环节，最后单独处理即可。

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#pragma comment(linker, "/stck:1024000000,1024000000")
#define lowbit(x) (x&(-x))
#define max(x,y) (x>=y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.1415926535897932384626433832
#define ios() ios::sync_with_stdio(true)
#define INF 0x3f3f3f3f
#define mem(a) ((a,0,sizeof(a)))
typedef long long ll;
ll t, a, b, c, d, v, T;
int main() {
    scanf("%lld", &T);
    while(T--) {
        scanf("%lld%lld%lld%lld%lld%lld", &a, &b, &c, &d, &v, &t);
        ll g = a/__gcd(a, c)*c;
        ll ans1=b+d-1,l=a,r=c;
        ll pos=0;
        while(l<g || r<g)
        {
            if(l<=r && l<g){
                if(l-pos>v) ans1+=(b-1);
                else ans1+=b;
                pos=l;
                l+=a;
            }
            else if(l>r && r<g){
                if(r-pos>v) ans1+=(d-1);
                else ans1+=d;
                pos=r;
                r+=c;
            }
        }
        if(l>=g) l-=a;
        if(r>=g) r-=c;
        ll mid=max(l,r);
        ans1 = t/g*ans1;
        if(g-mid<=v) ans1+=t/g;
        t = t%g;
        ll ans2=b+d-1;
        pos=0,l=a,r=c;
        while(l<=t || r<=t)
        {
            if(l<=r && l<=t){
                if(l-pos>v) ans2+=(b-1);
                else ans2+=b;
                pos=l;
                l+=a;
            }
            else if(l>r && r<=t){
                if(r-pos>v) ans2+=(d-1);
                else ans2+=d;
                pos=r;
                r+=c;
            }
        }
        printf("%lld\n",ans1+ans2);
    }
    return 0;
}