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 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 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 nn (1≤n≤1051≤n≤105) — the number of queries.

Next nn lines contain queries, one per line. Each line contains three integers pp, qq, and bb (0≤p≤10180≤p≤1018, 1≤q≤10181≤q≤1018, 2≤b≤10182≤b≤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

612=12=0,510612=12=0,510

43=1,(3)1043=1,(3)10

936=14=0,012936=14=0,012

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;
}

 

 

 

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