1020. Big Integer（高精度运算）

1. Big Integer
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Time Limit: 1sec Memory Limit:32MB
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
2

3
2 3 5
10

4
2 3 5 7
13
Sample Output
(0,1,0)
(1,1,3,6)

c语言

#include <stdio.h>
#include <string.h>
int mod(char dividend[], int divisor) {
    int p = 0;
    int temp = dividend[p++] - '0';
    while (1) {
        while (temp < divisor) {
            if (p == (strlen(dividend)) {
                return temp;
            } else {
                temp = temp * 10 + dividend[p++] - '0';
            }
        }
        temp %= divisor;
    }
}

int main() {
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        for (int i = 0; i < n; i++) {
        int m;
        int base[100];
        char bigint[400];
        scanf("%d", &m);
        for (int i = 0; i < m; i++) {
            scanf("%d", &base[i]);
        }
        scanf("%s", bigint);
        printf("(");
        for (int i = 0; i < m; i++) {
            printf("%d", mod(bigint, base[i]));
            if (i != m - 1) {
                printf(",");
            }

        }
            printf(")\n");
    }
    }
    return 0;
}

但是不幸超时了hahhh

cpp语言 可以正常通过。

#include<iostream>
#include<string>
using namespace std;
int mod(string dividend,int divisor)
{
    int p = 0;//对被除数的位数的指针
    int temp = dividend[p++] - '0';
    while(1)
    {
        while(temp < divisor)
        {
            if(p == dividend.size())    return temp;//如果指针超过被除数最低位还不能找到比除数小的数跳出，此时temp为答案
            temp = temp * 10 + dividend[p++] - '0';
        }
        temp %= divisor;
    }
}
int main()
{
    int t,n,b[102];
    string num;
    cin >> t;
    while(t--)
    {
        cin >> n;
        for(int i = 0;i < n;++i) cin >> b[i];
        cin >> num;
        cout << "(";
        for(int i = 0;i < n;++i)
        {
            cout << mod(num,b[i]);
            if(i != n - 1)  cout <<",";
        }
        cout << ")" << endl;
    }
    return 0;
}