链接：

题意:

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.

思路：

\((a*x+y)%b = (a%b*x%b+y%b)%b\)

代码：

#include<iostream>

#include<cstdio>

#include<cstring>

#include<string>

#include<algorithm>

#include<math.h>

#include<vector>

using namespace std;

typedef long long LL;

const int INF = 1e9;

const int MAXN = 1e6+10;

const int MOD = 1e9+7;

int main()

{

    int t, cnt = 0;

    LL n, x;

    scanf("%d", &t);

    while(t--)

    {

        scanf("%lld%lld", &n, &x);

        printf("Case %d: ", ++cnt);

        if (x%n == 0)

            puts("1");

        else

        {

            int cnt = 1;

            int tmp = x;

            while(x!=0)

            {

                x = (x*10+tmp)%n;

                cnt++;

            }

            printf("%d\n", cnt);

        }

    }

    return 0;

}

巴特西

LightOJ - 1078-Integer Divisibility（同余）

链接：

题意:

思路：

代码：

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