思路:
设输入的两个数分别为n和a,每一次所得到的数为update:
开始update=a,依次update分别为update*10+a,这样数据会超出范围,则update每次为update=(update*10+a)%n即可,
如果update=0,跳出循环;
只需证明:(update*10+a)%n=(update%n*10+a)%n即可;
由(update*10+a)%n=(update%n*10+a%n)%n,因为a<=n,所以a%n=a.证必;
1078 - Integer Divisibility
PDF (English) Statistics Forum Time Limit: 2 second(s) Memory Limit: 32 MB 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 Output for Sample Input 3 3 1 7 3 9901 1 Case 1: 3 Case 2: 6 Case 3: 12 PROBLEM SETTER: JANE ALAM JAN/******************************** author : Grant Yuan time : 2014/8/21 0:28 algorithm: Basic Math source : LightOj 1078**********************************/#includeusing namespace std;int t;int a,b,ans;int main(){ scanf("%d",&t); for(int i=1;i<=t;i++) { ans=1; scanf("%d%d",&a,&b); int temp=b; while(temp%a!=0) { temp=temp*10; temp+=b; temp%=a; ans++; } printf("Case %d: %d\n",i,ans); } return 0;}