Integer Divisibility（LightOJ-1078）

最新推荐文章于 2019-04-03 09:14:16 发布

Alex_McAvoy

最新推荐文章于 2019-04-03 09:14:16 发布

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分类专栏： # LightOJ # 数论——整除与同余

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数论——整除与同余

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本文探讨了一种使用同余定理解决数学问题的方法，具体为找到一个整数n的倍数，该倍数仅由指定数字构成，并确定这个倍数的位数。通过实例解释了算法思路，提供了源代码实现。

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Problem 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.

Examples

Input

3
3 1
7 3
9901 1

Output

Case 1: 3
Case 2: 6
Case 3: 12

题意：t 组数据，每组给出两个数 n 和 m，让 m 变为 mm、mmm...，问当 m 变为几位数时恰好能整除 n

思路：同余定理

实际是考察同余定理，对于两个数 A、B，有：

(A + B) mod M = ( A mod M + B mod M ) mod M
(A * B) mod M = ((A mod M) *( B mod M)) mod M

因此，令 A = m，每次让 A*10 后再加 m 然后 mod n，直到能整除为止

Source Program

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<deque>
#include<vector>
#include<set>
#include<map>
#define PI acos(-1.0)
#define E 1e-6
#define INF 0x3f3f3f3f
#define N 100001
#define LL long long
const int MOD=1e9+7;
const int dx[]={-1,1,0,0};
const int dy[]={0,0,-1,1};
using namespace std;

int main(){
    int t;
    scanf("%d",&t);
    int Case=1;
    while(t--){
        LL n,m;
        scanf("%lld%lld",&n,&m);

        int temp=m;
        int res=1;
        while(temp%n){
            temp=(temp*10+m)%n;
            res++;
        }

        printf("Case %d: %lld\n",Case++,res);
    }
    return 0;
}