B. Position in Fraction
time limit per test
1 secondmemory limit per test
256 megabytesinput
standard inputoutput
standard outputYou have a fraction . You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers a, b, c (1 ≤ a < b ≤ 105, 0 ≤ c ≤ 9).
Output
Print position of the first occurrence of digit c into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.
Examples
input
1 2 0
output
2
input
2 3 7
output
-1
Note
The fraction in the first example has the following decimal notation: . The first zero stands on second position.
The fraction in the second example has the following decimal notation: . There is no digit 7 in decimal notation of the fraction.
【题意】:给定分子m 分母n和一个数c,找出c在m/n的循环节第几个位置出现,没出现过输出-1。
【分析】:模拟实现循环节,记录出现过的下标。
【代码】:
#include <bits/stdc++.h> using namespace std; typedef long long LL; typedef unsigned long long ULL; const int maxn = 1e5+1000; int i,pos=-1,c; int main() { int n,m,i; scanf("%d%d%d",&m,&n,&c); i=0; int cnt=0; int fz=m,fm=n; while(1) { i++; fz=fz*10; //printf("%d",fz/fm); if(fz/fm==c) { pos=i; break; } fz=fz%fm; if(i>=100005) break; //限制循环节长度 } printf("%d ",pos); return 0; }