思路：

??? 設(shè)輸入的兩個數(shù)分別為n和a,每一次所得到的數(shù)為update：

??? 開始update=a,依次update分別為update*10+a,這樣數(shù)據(jù)會超出范圍，則update每次為update=(update*10+a)%n即可，

??? 如果update=0,跳出循環(huán)；

??? 只需證明：(update*10+a)%n=(update%n*10+a)%n即可；

????????????????????? 由(update*10+a)%n=(update%n*10+a%n)%n，因為a<=n，所以a%n=a.證必；

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 Yuantime : 2014/8/21 0:28algorithm: Basic Mathsource : LightOj 1078 **********************************/ #include<bits/stdc++.h>using 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; }

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轉(zhuǎn)載于:https://www.cnblogs.com/codeyuan/p/4254449.html

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