Position in Fraction （小数除法模拟）

最新推荐文章于 2019-11-20 19:22:30 发布

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最新推荐文章于 2019-11-20 19:22:30 发布

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分类专栏： # 数论

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本文介绍了一种算法，用于寻找特定数字在给定分数的小数展开中首次出现的位置。通过不断将分子乘以10并与分母进行除法运算来实现这一目标。如果找到指定数字，则返回其位置；若循环节过长仍未找到则返回-1。

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Position in Fraction

You have a fraction $\text{[math]}$ . 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

Input

2 3 7

Output

-1

Note

The fraction in the first example has the following decimal notation: $\text{[math]}$ . The first zero stands on second position.

The fraction in the second example has the following decimal notation: $\text{[math]}$ . There is no digit 7 in decimal notation of the fraction.

code：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int main(){
    int n,m,c,i;
    scanf("%d%d%d",&m,&n,&c);
    int pos = -1;
    int fz = m,fm = n;
    i = 0;
    while(1){
        i++;
        fz *= 10;
        if(fz / fm == c){
            pos = i;
            break;
        }
        fz = fz % fm;
        if(i >= 100005) break;//限制循环节长度
    }
    printf("%d\n",pos);
    return 0;
}