豎式除法模擬

寫這篇博客的目的是為了更好利用數學方法來處理數據，之前遇到了一道題給你a,b,c三個數問a/b小數點后幾位是c，按照我的想法肯定是現將a/b的結果使用函數轉換成字符串，然后直接在字符串中查找，很遺憾對於數據量較大的數根本就不能通過，還好這道水題是我隊友AC通過了，這道題到我手里肯定是通不過的，那我們就來模擬一下豎式除法的過程。

 

看到這個式子是不是回憶起小學學習除法的經歷了吧，這應該是四則運算中最困難的了。

我們來看一下運算過程:

被除數a對除數b整除，得到一個結果，再對余數乘10（十進制），組成新余數，新余數變為新被除數，除數不變，再次進行整除，不斷循環。

也就是a=a%b*10

 

例題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

 

題目意思：問1/n后的第m位數是多少。

#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\n",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.

 

題目意思：問a/b得到的小數中c在小數點后多少位上。

 

#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;///若沒有找到，即為初始化的-1
    for(i=1; i<MAX; i++)
    {
        a=a*10;
        if(a/b==c)
        {
            ans=i;
            break;
        }
        a=a%b;
    }
    printf("%d\n",ans);
    return 0;
}