[sicily online]1020. Big Integer

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 = 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

232 3 51042 3 5 713

Sample Output

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

題目分析：

第一次做時，本來想做個字串類比整型資料的加、減、乘、除、模運算，用了自己的庫，所以把題目中的大整數和除數都用string來表示，然後用自己寫的string模數，最後總是逾時，看了別的大神的做法，原來除數沒必要用string表示（因為小於1000），這樣就不會逾時了

#include<iostream>#include<stdio.h>#include<cmath>#include<iomanip>#include <vector>#include <string>#include <algorithm>#include <sstream>using namespace std;string strMod(string x1,int x2){int num=0;for(string::size_type i=0;i<x1.size();i++){num=num*10+(x1[i]-'0');num=num%x2;}//直接把int轉化成stringstringstream strStream;strStream<<num;string result=strStream.str();return result;}int main(){int n;cin>>n;for(int i=0;i<n;i++){int m;scanf("%d",&m);vector<int> data;int tmpStr;for(int j=0;j<m;j++){cin>>tmpStr;data.push_back(tmpStr);}//end forstring bigInteger;cin>>bigInteger;cout<<"(";for (int j=0;j<m-1;j++){string tmp=strMod(bigInteger,data[j]);printf("%s,",tmp.c_str());}string tmp=strMod(bigInteger,data[m-1]);printf("%s)\n",tmp.c_str());}}