ppt內容
大數取模
一個大數對一個數取餘，可以把大數看成各位數的權值與個
位數乘積的和。
比如1234 = ((1 * 10 + 2) * 10 + 3) * 10 + 4，對這個數進
行取餘運算就是上面基本加和乘的應用。
代碼實現
int len = a.length();
int ans = 0;
for(int i = 0; i < len; i++){
ans = (ans * 10 + a[i] - ’0’) mod b;
}
E - 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.
Sample Input
3
3 1
7 3
9901 1
Sample Output
Case 1: 3
Case 2: 6
Case 3: 12
題意：給定兩個數n,d，d是一位數字，問最小的能整除n且只含數字d的數是幾位數。
思路：從後先前分析：隨意給定一個只含d的數，這個數可以分爲兩部分的和：一部分是能被n整除的，另一部分對n的餘數。如果這個數不能被n整除，那麼這個數就要再多一位了，就等於把現在的數*10+d,現在能被n整除的部分乘10以後依然能被n整除，那麼我們就只需考慮現餘數的部分了，只要現在餘數的部分乘以10+d能被n整除就找到了這個數
代碼：
#include <stdio.h>
typedef long long ll;//要用long long型
int main()
{
ll t,k=0;
scanf("%lld",&t);
while(t--)
{
ll n,d,z=1,i;//z初值等於1
k++;
scanf("%lld%lld",&n,&d);
ll r=d;
while(r%n)
{
z++;
r=r%n*10+d;//改變的是r，d是一個定值
}
printf("Case %lld: %lld\n",k,z);
}
return 0;
}