竖式除法模拟

写这篇博客的目的是为了更好利用数学方法来处理数据，之前遇到了一道题给你a,b,c三个数问a/b小数点后几位是c，按照我的想法肯定是现将a/b的结果使用函数转换成字符串，然后直接在字符串中查找，很遗憾对于数据量较大的数根本就不能通过，还好这道水题是我队友AC通过了，这道题到我手里肯定是通不过的，那我们就来模拟一下竖式除法的过程。

看到这个式子是不是回忆起小学学习除法的经历了吧，这应该是四则运算中最困难的了。

我们来看一下运算过程:

被除数a对除数b整除，得到一个结果，再对余数乘10（十进制），组成新余数，新余数变为新被除数，除数不变，再次进行整除，不断循环。

也就是a=a%b*10

例题1：HDU 2117  http://acm.hdu.edu.cn/showproblem.php?pid=2117

Just a Numble

Problem Description

Now give you two integers n m, you just tell me the m-th number after radix point in 1/n,for example n=4,the first numble after point is 2,the second is 5,and all 0 followed

 

Input

Each line of input will contain a pair of integers for n and m(1<=n<=10^7,1<=m<=10^5)

 

Output

For each line of input, your program should print a numble on a line,according to the above rules

 

Sample Input

4 2

5 7

123 123

 

Sample Output

5

0

8

 

题目意思：问1/n后的第m位数是多少。

#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
    int m,n,result,i;
    int temp;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        temp=1;
        for(i=1;i<=m;i++)
        {
            temp*=10;
            result=temp/n;
            temp=temp%n;
        }
        result=result%10;
        printf("%d
",result);
    }
    return 0;
}

例2

CF 900B - Position in Fraction

Description

You 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.

Sample Input

Input

1 2 0

Output

2

Input

2 3 7

Output

-1

Hint

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.

题目意思：问a/b得到的小数中c在小数点后多少位上。

#include <iostream>
#include<cstdio>
#include<algorithm>
const int MAX=1e6+10;
using namespace std;
int main()
{
    int a,b,c,i,ans;
    scanf("%d%d%d",&a,&b,&c);
    ans=-1;///若没有找到，即为初始化的-1
    for(i=1; i<MAX; i++)
    {
        a=a*10;
        if(a/b==c)
        {
            ans=i;
            break;
        }
        a=a%b;
    }
    printf("%d
",ans);
    return 0;
}