sicily 1020. Big Integer

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 = b1*b2*...*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

Copy sample input to clipboard

Sample Output

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

#include <iostream>
#include <string>
#include <string.h>
using namespace std;
             
int* findMod(string str, int* arr, int len) {
    int size = str.size();
    int* arrt = new int[len];
    memset(arrt, 0, sizeof(int) * len);
             
    for (int i = 0; i != size; ++i) {
       for (int j = 0; j != len; ++j) {
         arrt[j] = (arrt[j] * 10 + (str[i] - '0')) % arr[j]; 
       }   
    }        
    return arrt;
}            
             
int main(int argc, char* argv[])
{            int T, n, *arr;
    string x;
    cin >> T;
    while (T--) {
       cin >> n;
       arr = new int[n];
       for (int i = 0; i != n; ++i)
          cin >> arr[i];
       cin >> x;
       int *result = findMod(x, arr, n); 
       cout << "(";
       for (int i = 0; i != n - 1; i++)
          cout << result[i] << ",";
       cout << result[n - 1] << ")" << endl;
    }      
             
    return 0;
}

因为给的空间还是很大的，所以我在mod的时候用空间换时间，其实这样实现是不好的，因为申请的空间根本没释放。下面这样的话比较好一点，时间上也只是差了0.07多