CodeforcesRound#483(Div.2)C.Finiteornot?
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 integerspp,qqandbb. You need to answer whether the result ofp/qp/qin notation with basebbis a finite fraction.
A fraction in notation with basebbis 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 integernn(1≤n≤1051≤n≤105)— the number of queries.
Nextnnlines contain queries, one per line. Each line contains three integerspp,qq, andbb(0≤p≤10180≤p≤1018,1≤q≤10181≤q≤1018,2≤b≤10182≤b≤1018). All numbers are given in notation with base1010.
Output
For each question, in a separate line, printFiniteif the fraction is finite andInfiniteotherwise.
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
题意：p/q能否表示为b进制下的有限小数形式,即1/q能否表示为b进制下的有限小数形式，类似十进制小数转二进制，
如0.125，小数乘以进制取整数0，小数0.25乘以进制取整数0，小数0.5乘以进制取整数1，且乘到1停止
0.125对应 0.001 即0*（1/2）+0*（1/4）+1*（1/8）=0.125，回看过程，1/q*b*b....每次取整数，最后等于1，相当于乘以
b的过程是对q约分，那么只要b包含q全部的因子即可。
#include <iostream>
#define ll long long using namespace std; ll gcd(ll a,ll b) { if(!b)return a; return gcd(b,a%b); } int main() { cin.sync_with_stdio(0); cin.tie(0); cout.tie(0); int n; cin>>n; while(n--) { ll p,q,b; cin>>p>>q>>b; ll g=gcd(p,q); p/=g; q/=g;//约分，将分母化成最简。 if(p==0||q==1)//p==0 该数值为0显然任何形式都可以表示； q==1分母为1表示整数 ，都可以表示 { cout<<"Finite"<<endl; continue; } //若1/q可以被表示，则p/q也能被表示 while(q!=1&&b!=1) { b=gcd(b,q); q/=b; //约分 ，b必须含有q的全部因子才是有限小数 } if(q==1)printf("Finite "); else printf("Infinite "); } return 0; }
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