CF#483(div2 C)

http://codeforces.com/contest/984/problem/C

C. Finite or not

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given several queries. Each query consists of three integers $\mathrm{pp, qq and bb.\; You\; need\; to\; answer\; whether\; the\; result\; of p/qp/q in\; notation\; with\; base bb is\; a\; finite\; fraction.}$

A fraction in notation with base $\mathrm{bb is\; finite\; if\; it\; contains\; finite\; number\; of\; numerals\; after\; the\; decimal\; point.\; It\; is\; also\; possible\; that\; a\; fraction\; has\; zero\; numerals\; after\; the\; decimal\; point.}$

Input

The first line contains a single integer $\mathrm{nn (1\le n\le 1051\le n\le 105) —\; the\; number\; of\; queries.}$

Next $\mathrm{nn lines\; contain\; queries,\; one\; per\; line.\; Each\; line\; contains\; three\; integers pp, qq,\; and bb (0\le p\le 10180\le p\le 1018, 1\le q\le 10181\le q\le 1018, 2\le b\le 10182\le b\le 1018).\; All\; numbers\; are\; given\; in\; notation\; with\; base 1010.}$

Output

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

Examples

input

Copy

2
6 12 10
4 3 10

output

Copy

Finite
Infinite

input

Copy

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

output

Copy

Finite
Finite
Finite
Infinite

Note

$\frac{\mathrm{612=12=0,510612=12=0,510}}{}$

$\frac{\mathrm{43=1,(3)1043=1,(3)10}}{}$

$\frac{\mathrm{936=14=0,012936=14=0,012}}{}$

$\frac{\mathrm{412=13=0,13412=13=0,13}}{}$

题目标签：数论；

题目大意：判断p/q在b进制下是否为有限小数；

0.0

题目思路：不断的用b去消除q中的因子，但是求出一个g=gcd（q，b）后，必须直接将q中的所有g都除尽，否则会超时；标程是不断的寻找b=gcd（q，b），因为每次q除尽最大公约数后，q中含有的只能是原来的最大公约数中的因子，所以不用每次用b去更新q，可以每次更新完q后，也更新b值;

小数的二进制转换：乘基取整，顺序排列；

#include <iostream>
#include <cstdio>
//小数的二进制转换：乘基取整，顺序排列；
//b的k次方%q==0;
//判断k个b能否将q的所有因子均消耗完;
using namespace std;
typedef long long ll;

ll gcd(ll x,ll y)
{
    if(y==0)return x;
    return gcd(y,x%y);
}
int main()
{
    int n;
    ll p,q,b;
    scanf("%d",&n);
    while(n--)
    {
        bool flag=false;
        scanf("%I64d%I64d%I64d",&p,&q,&b);
        if(p==0)q=1;
        q/=gcd(p,q);
        ll g=gcd(q,b);
        while(g!=1){        //q,b互质；
            while(q%g==0)q/=g;
            g=gcd(q,b);
        }
        if(q==1)
            printf("Finite\n");
        else
            printf("Infinite\n");
    }
    return 0;
}