Light--1078(取模运算)
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Description
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
解题思路:一开始以为是什么找规律题,找了2个小时,愣是没什么收获,后来才知道,是一道简单的取模运算.但注意题n不一定大于digit.
代码如下:
#include<stdio.h>int main(){long long n,cas,k,m,s;cas=1; int t; scanf("%d",&t); while(t--){ m=1; scanf("%lld%lld",&n,&k); s=k; printf("Case %lld: ",cas++); if(k%n==0){ printf("1\n"); continue; } while(k!=0){ k=(k*10+s)%n; m++; } printf("%lld\n",m); }return 0;}
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