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LightOj 1078 Basic Math
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 **********************************/ #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; }