lightoj-1078-Integer Divisibility【大数除法】

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1078 - Integer Divisibility

If an integer is not divisible by 2 or 5, some multiple of that number in decimal notation is a sequence of only a digit. Now you are given the number and the only allowable digit, you should report the number of digits of such multiple.

For example you have to find a multiple of 3 which contains only 1's. Then the result is 3 because is 111 (3-digit) divisible by 3. Similarly if you are finding some multiple of 7 which contains only 3's then, the result is 6, because 333333 is divisible by 7.

Input

Input starts with an integer T (≤ 300), denoting the number of test cases.

Each case will contain two integers n (0 < n ≤ 10⁶ and n will not be divisible by 2 or 5) and the allowable digit (1 ≤ digit ≤ 9).

Output

For each case, print the case number and the number of digits of such multiple. If several solutions are there; report the minimum one.

一直对大数掌握不好，要多拍这类题才行，就是模拟除法的列式计算

代码1：

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
int n,m,base,ans;
int main()
{
  int t,text=0;
  scanf("%d",&t);
  while(t--)
  {
    ans=1;
    scanf("%d %d",&n,&m);
    base=m;
    while(m%n!=0)
    {
      m=m%n;
      m=m*10+base;
      ans++;
    }
    printf("Case %d: %d\n",++text,ans);
  }
  return 0;
}

代码2：

#include<cstdio>
#include<algorithm>
using namespace std;
int n,m,cnt;
int main()
{
  int i,t,text=0;
  scanf("%d",&t);
  while(t--)
  {
    scanf("%d %d",&n,&m);
    cnt=m;
    for(i=1;;i++)
    {
      if(cnt%n==0)  break;
      cnt=(cnt*10+m)%n;
    }
    printf("Case %d: %d\n",++text,i);
  }
  return 0;
}