Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

We suspect that for every positive integer N there exists an integer of the form 11...10...0 (a sequence of 1's followed by 0 or more 0's) that is divisible by N . For example, with N = 3 , 111 is divisible by 3, with N = 4 , 100 is divisible by 4, with N = 7 , 11111 is divisible by 7. We want to verify this for some integers. The solution to this problem is to find two different numbers P and Q in the form of 11...1 (a sequence of 1's) that have the same remainder when dividing by N . The difference D between P and Q will be in the form of 11...10...0 and divisible by N .

In order to solve this problem, we have to start with finding the remainder when dividing a number in the form of 11...1 by N . Your task is to write a program to do this.

Input 

The input file consists of several data sets. The first line of the input file contains the number of data sets which is a positive integer and is not bigger than 20. The following lines describe the data sets.

Each data set is described by two lines. The first line contains the integer N (1 < N < 10⁹) . The second line contains the integer number P (P contains at least one digit and at most 2000 digits).

Output 

For each test case, write in one line the remainder when dividing P by N .

Sample Input 

2
4 
11 
5 
111

Sample Output 

3 
1

---

#include <stdio.h>

int main() {
    int t, n, i;
    char s[3000];
    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);
        scanf("%s", s);
        int tmp = 0;
        for(i = 0; s[i]; i++) {
            tmp = tmp*10 + s[i]-'0';
            tmp %= n;
        }
        printf("%d\n", tmp);
    }
    return 0;
}

---

import java.math.BigInteger;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        BigInteger a, b;
        int t;
        t = cin.nextInt();
        while(t-- != 0) {
            a = new BigInteger(cin.next());
            b = new BigInteger(cin.next());
            System.out.println(b.mod(a).toString());
        }
    }
}