CodeForces 900B Position in Fraction_codeforces position in fraction-优快云博客

本文链接：https://blog.youkuaiyun.com/CS33sun/article/details/78992423

本文介绍了一个简单的程序，用于解决一个特定的问题：给定三个整数a、b和c，找到分数a/b的小数点后第一次出现数字c的位置。如果该数字不存在，则返回-1。通过直接模拟的方法解决了这个问题，程序使用C语言编写。

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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.

Example

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.

题意：多组数据，

每组数据输入三个数 a,b,c

输出，a/b的第几位数是c

很水的吧，直接模拟就行了，数学上好像有证明，不会有太多位的，所以大胆写吧！

#include <stdio.h>
int main() {  
    int a, b, c;  
    while(~scanf("%d%d%d",&a,&b,&c)) {
        for(int i = 1; i < 255; i++) {
            a *= 10;
            if(a/b == c) {
            	printf("%d\n",i);
				break;
			}
            a %= b;
			if(i == 254) printf("-1\n");  
        }
    }   
    return 0;  
}