Member Avatar for rebelvarun
rebelvarun 0 Newbie Poster

the question is from ----acm.timus.ru "PROBLEM 1095"-VOLUME 1-"Nikifor 3"
Nikifor knows that a certain positive integer has in its decimal form each of the digits 1,2,3,4. You are asked to determine if Nikifor can rearrange the digits of the number in such a way that the new number divides by 7.
Input:

The first line of the input contains the number N (not exceeding 10000) of positive integers that are to be checked. The next N lines contain these integers. Each number has no more than 20 digits.
Output:

For each of the N numbers you should output a number divisible by 7 that can be obtained from the corresponding number from the input file by a rearrangement of the digits. If such rearrangement does not exist you should output 0 in the corresponding line. In the case of several valid rearrangements you may find only one of them.
Kindly look in to the attached file and tell the flaw in the code....:-)....->

#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
#include<math.h>
using namespace std;
long long int convert(string s);
int main()
{
	int n;
	cin>>n;
	string num;
	int p1,p2,p3,flag=0;
	long long int N=0;
	int rem,var;
	char ch;
	vector<string>v;
	string s="1234";
	string t;
	while(n--){
		cin>>num;
		p1=num.find_first_not_of('0');
		num.erase(0,p1);
		num=" "+num+" ";
		p1=num.find_first_of('1');
		num.erase(p1,1);
		p1=num.find_first_of('2');
		num.erase(p1,1);
		p1=num.find_first_of('3');
		num.erase(p1,1);
		p1=num.find_first_of('4');
		num.erase(p1,1);
		num.erase(0,1);
		num.erase(num.length()-1,1);
		p1=num.find_first_of('0');
		p2=num.find_first_not_of('0');
		if(p1==0&&p2<0){
			num="3241"+num;
			v.push_back(num);
			continue;
		}
		if(p1==0&&p2>0){
			ch=num.at(p2);
			num.erase(p2,1);
			num=ch+num;
			cout<<num<<endl;
		}	
		N=convert(num);
		rem=N%7;
		ch=rem+48;
		do{
			flag=0;
			t=ch+s;
			var=convert(t);
			if(var%7==0){
				num=num+s;
				v.push_back(num);
				flag=1;
				break;
			}
		}while(next_permutation(s.begin(),s.end()));
		if(flag==0)
			v.push_back("0");
	}
	vector<string>::iterator itr;
	for(itr=v.begin();itr!=v.end();itr++)
		cout<<*itr<<endl;
		
}	
long long int convert(string num){
	long long int N=0;
	for(int i=0;i<num.length();i++){
		N=N+(num.at(i)-48)*pow(10,num.length()-1-i);
	}
	return N;
}
  • 1 Contributor
  • 0 Replies
  • 52 Views
Be a part of the DaniWeb community

We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge.