#include<iostream.h>
#include<string.h>
int main()
{
        int flag=0,i=0,j=0;
        char ar[4]="abc" ;
        char temp[4];
        cout<<"Plz enter a string with size #3";
        cin>>temp;
for(j=0;ar[j];j++)
        {
                flag=0;
                for(i=0;temp[i];i++)
                {
                        if(ar[j]==temp[i])
                                flag+=1;
                }
                if(flag==0||flag>1)
                        break;
        }
        if(flag==0||flag>1)
                cout<<"Not accepted";
        else
		cout<<"Accepted";
}
Note:PLz convert "iostream.h" into "stdio.h" and "cout<<" 
to "printf"

The above is broken.  If the ar[4] is "aac" rather 
than "abc" it fails.

#include "stdio.h"

#define CountOf(a) (sizeof(a)/sizeof(*a))

int main(int argc, _TCHAR* argv[])
{


	char data[] = "abc";
	char test[CountOf(data)];

	printf("Enter %d characters: ", CountOf(data)-1);
	scanf("%s",  test);

	for (int i=0;i < CountOf(data);i++)
	{
		for (int j=0;j < CountOf(data);j++)
		{
			if (data[i] == test[j]) 
			{
				test[j] = 0; 
				break;
			}
		}
	}

	for (int i=0;i < CountOf(data);i++)
	{

		if (test[i]) {
			printf("\nNot ");
			break;
		}
	}
	printf("\nAccepted");

	return 1;
}

both of the solutions above given have complexity O(N^2),
i have tried to solve it in O(N).let me inform if there is 
any bug or better solution .

#include<iostream>
using namespace std;
int main()
{
    int a[130],a1[130];
    memset(a,0,sizeof(a));
    memset(a1,0,sizeof(a));
    char ans[1000];
    char str[]="putanystringhere";
    for(int i=0;i<sizeof(str)/sizeof(char)-1;i++)
    {
        a[(int)str[i]]+=1; 
    }
    cout<<"Enter any string of length
"<<sizeof(str)/sizeof(char)-1<<" :";
    cin>> ans;  
    for(int i=0;i<sizeof(str)/sizeof(char)-1;i++)
    {
        a1[(int)ans[i]]+=1; 
    }
    bool flag=true;
    for(int i=0;i<sizeof(str)/sizeof(char)-1;i++)
    {
           if(a[(int)str[i]]!=a1[(int)str[i]])
            {
                     flag=false;
                     break;
            }
    }
    if(flag==true)
    cout<<"No error";
    else
    cout<<"Error";
    system("pause");
    return 0;
}

My answer is O(n) - but it is done in C# sorry for C guys
        public bool isAccepted(string original, string
toCompare)
        {
            if (original.Length != toCompare.Length)
                return false;
            bool[] flags = new bool[128];
            for (int i = 0; i < toCompare.Length; i++)
                flags[toCompare[i]] = true;
            for (int i = 0; i < original.Length; i++)
                if (!flags[original[i]])
                    return false;
            return true;
        }

if we ask the user to enter the second string having the 
same characters as in the first string, the program 
essentailly reduces to finding whether there is a 
repetition in the second string. A program for this is 
given below.

#include<iostream>
using namespace std;

int main()
{
	char string1[10];
	char string2[10];
	int flag = 0;

	cout<<"\n Enter the first string";
	cin>>string1;
	cout<<"\n Enter the second string with the same 
characters as entered in string1";
	cin>>string2;
	for(int i = 0; i < strlen(string2); i++)
	{
		for(int j = i+1; j < strlen(string2) ; j++)
		{
			if(string2[i] == string2[j])
			{
				flag = 1;
				break;
			}
		}
	}

	if(flag ==1)
		cout<<"\n wrong";
	else
		cout<<"\n correct";

	return 0;
}