两数相除,判断小数位是否有限位
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You are given several queries. Each query consists of three integers p, q and b. You need to answer whether the result of p/q in notation with base b
is a finite fraction.
A fraction in notation with base b
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 n
(1≤n≤105
) — the number of queries.
Next n
lines contain queries, one per line. Each line contains three integers p, q, and b (0≤p≤1018, 1≤q≤1018, 2≤b≤1018). All numbers are given in notation with base 10
.
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=0,510
43=1,(3)10
936=14=0,012
412=13=0,13
题意 : 给两个数,以及一个所要求的进制,询问是否在此进制下,相除后是否为有限位小数
思路分析 : 比赛的时候想了一个模拟,不过没有去写,旁边的大佬一直再和说,会超时,会超时,然后赛后才知道是用这样的,比如 10进制下,判断两数是否整除,只需要让除数已知除以 2 或者 5,当可以除到 1 的时候,代表是可以整除的,如果是换成任意进制下的一个数,则需要去除当前的数和 b进制的约数。
代码示例 :
#define ll long long ll gcd(ll a, ll b){ return b == 0?a:gcd(b, a%b); } int main() { //freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); ll t; ll p, q, b; cin >> t; while(t--){ scanf("%lld%lld%lld", &p, &q, &b); ll g = gcd(p, q); ll x = q / g; ll f = gcd(b, x); while(1){ if (f == 1) break; while(x%f == 0) x /= f; f = gcd(b, x); } if (x == 1) printf("Finite\n"); else printf("Infinite\n"); } return 0; }
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