sicily_1020 Big Integer

题目

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

Long long ago, there was a super computer that could deal with VeryLongIntegers(no VeryLongInteger will be negative). Do you know how this computer stores the VeryLongIntegers? This computer has a set of n positive integers: b1,b2,...,bn, which is called a basis for the computer.

The basis satisfies two properties:

1. 1 < bi <= 1000 (1 <= i <= n),
2. gcd(bi,bj) = 1 (1 <= i,j <= n, i ≠ j).

Let M = b1b2...*bn

Given an integer x, which is nonegative and less than M, the ordered n-tuples (x mod b1, x mod b2, ..., x mod bn), which is called the representation of x, will be put into the computer.

Input

The input consists of T test cases. The number of test cases (T) is given in the first line of the input.
Each test case contains three lines.
The first line contains an integer n(<=100).
The second line contains n integers: b1,b2,...,bn, which is the basis of the computer.
The third line contains a single VeryLongInteger x.

Each VeryLongInteger will be 400 or fewer characters in length, and will only contain digits (no VeryLongInteger will be negative).

Output

For each test case, print exactly one line -- the representation of x.
The output format is:(r1,r2,...,rn)

Sample Input

2

3
2 3 5
10

4
2 3 5 7
13

Sample Output

(0,1,0)
(1,1,3,6)

题目大意

给定一个高精度数，求出它除以n个整形数的模，并写成向量的形式。

思路

在处理数的过程中只留下模，防止数据溢出。

代码

#include
#include
using std::cin;
using std::cout;
using std::string;

int mod(string divided, int divisor) {
  int p = 0, tmp = 0;
  tmp = divided[p++] - '0';
  while (true) {
    while (tmp < divisor) {
      if (p == divided.length()) return tmp;
      tmp = 10 * tmp + divided[p++] - '0';
    }
    tmp %= divisor;
  }
}

int main() {
  int T;

  cin >> T;
  while (T--) {
    int n, b[100 + 1];
    string x;

    cin >> n;
    for (int i = 0; i < n; i++) cin >> b[i];
    cin >> x;

    cout << '(';
    for (int Head = 1, i = 0; i < n; Head = 0, i++) {
      if (!Head) cout << ',';
      cout << mod(x, b[i]);
    }
    cout << ')' << endl;
  }
  return 0;
}