You are given several queries. Each query consists of three integers $\mathrm{p,\; q\; and\; b.\; You\; need\; to\; answer\; whether\; the\; result\; of\; p/q\; in\; notation\; with\; base\; b}$

is a finite fraction.

A fraction in notation with base $b$

is finite if it contains finite number of numerals after the decimal point. It is also possible that a fraction has zero numerals after the decimal point.

The first line contains a single integer $n$

$\mathrm{(1\le n\le 105}$

) — the number of queries.

Next $n$

$\mathrm{lines\; contain\; queries,\; one\; per\; line.\; Each\; line\; contains\; three\; integers\; p,\; q,\; and\; b\; (0\le p\le 1018,\; 1\le q\le 1018,\; 2\le b\le 1018).\; All\; numbers\; are\; given\; in\; notation\; with\; base\; 10}$

.

For each question, in a separate line, print Finite if the fraction is finite and Infinite otherwise.

2
6 12 10
4 3 10

Finite
Infinite

4
1 1 2
9 36 2
4 12 3
3 5 4

Finite
Finite
Finite
Infinite

$\frac{\mathrm{612=12=0,510}}{}$

$\frac{\mathrm{43=1,(3)10}}{}$

$\frac{\mathrm{936=14=0,012}}{}$

$\frac{\mathrm{412=13=0,13}}{}$

題意 ： 給兩個數，以及一個所要求的進制，詢問是否在此進制下，相除後是否為有限位小數

思路分析 ： 比賽的時候想了一個模擬，不過沒有去寫，旁邊的大佬一直再和說，會超時，會超時，然後賽後才知道是用這樣的，比如 10進制下，判斷兩數是否整除，只需要讓除數已知除以 2 或者 5，當可以除到 1 的時候，代表是可以整除的，如果是換成任意進制下的一個數，則需要去除當前的數和 b進制的約數。

代碼示例 ：

#define ll long long

ll gcd(ll a, ll b){
    return b == 0?a:gcd(b, a%b);
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    ll t;
    ll p, q, b;
    
    cin >> t;
    while(t--){
        scanf("%lld%lld%lld", &p, &q, &b);
        ll g = gcd(p, q);
        ll x = q / g;
        ll f = gcd(b, x);
        
        while(1){
            if (f == 1) break;
            while(x%f == 0) x /= f;
            f = gcd(b, x);
        }   
        if (x == 1) printf("Finite\n");
        else printf("Infinite\n");
    }
    return 0;
}

兩數相除，判斷小數位是否有限位