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.
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.
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
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
题意:给出p,q,b,询问在b进制下,p/q是否为无限小数。
题解:首先想到的是在十进制下p/q何时为无限小数,发现当分式最简的时候,分母的因子只存在10的质因子时,一定不是无限小数,反之则是无限小数,则推广到b,将q中所有b的因子除掉,看是否可以将q完全除掉即可。
#include<stdio.h> #include<algorithm> #include<iostream> #include<string.h> #include<vector> #include<stdlib.h> #include<math.h> #include<queue> #include<deque> #include<ctype.h> #include<map> #include<set> #include<stack> #include<string> #include<algorithm> #define gcd(a,b) __gcd(a,b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define FAST_IO ios::sync_with_stdio(false) #define mem(a,b) memset(a,b,sizeof(a)) const double PI = acos(-1.0); const double eps = 1e-6; const int MAX=1e5+10; const long long INF=0x7FFFFFFFFFFFFFFFLL; const int inf=0x3f3f3f3f; const int mod=1e9+7; typedef long long ll; using namespace std; int main() { int t; cin>>t; while(t--) { ll p,q,b; scanf("%I64d %I64d %I64d",&p,&q,&b); ll GCD=b; q/=gcd(p,q); while(q!=1) { GCD=gcd(GCD,q); if(GCD==1) break; q/=GCD; } if(q==1) printf("Finite\n"); else printf("Infinite\n"); } return 0; }