Codeforces Round #483 (Div. 2) C. Finite or not?

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

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

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

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

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

 

 这题，看这个分数在b进制下是否为无限循环小数。

一个分数是否为无限循环小数取决于分母。

于是想到这题一定是考虑 分母q 和 进制b 的关系，

于是就是想办法  把分母化为1就不是无限循环小数

temp1 = gcd(q, b);

if (temp1 == 1) break; while(q % temp1== 0) { q /= temp1; }
while一定要加 不然会超时。

 

#include<bits/stdc++.h>

using namespace std;
const int maxn = 3e5 + 10;
typedef long long LL;
LL gcd(LL a, LL b ) {
    if (b == 0) return a;
    return gcd(b, a % b);
}
int main() {
    int n;
    LL p, q, b;
    scanf("%d", &n);
    while(n--) {
        scanf("%lld%lld%lld", &p, &q, &b);
        LL g = gcd(p, q);
        p = p / g, q = q / g;
        LL temp = p / q;
        p = p - temp * q;
        if (!p) {
            printf("Finite\n");
            continue;
        }
        LL temp1;
        while(1) {
            temp1 = gcd(q, b);
            if (temp1 == 1) break;
            while(q % temp1== 0) {
                q /= temp1;
            }
        }
        if (q == 1 ) printf("Finite\n");
        else printf("Infinite\n");
    }
    return 0;
}