lightoj 1078 - Integer Divisibility 【同余定理】
2015-12-03 14:27
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1078 - 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.
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).
同余定理:(a+b)%n==(a%n+b%n)%n;
代码:
PDF (English) | Statistics | Forum |
Time Limit: 2 second(s) | Memory Limit: 32 MB |
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 |
代码:
#include<stdio.h> #include<string.h> int main() { int T; int n,m; int c=1; scanf("%d",&T); while(T--) { scanf("%d%d",&n,&m); int a=m%n; int k=1; while(a) { a=(a*10%n+m%n)%n; k++; } printf("Case %d: %d\n",c++,k); } return 0; }
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