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Monday, May 26, 2014

NBIN-New Binary

Below given code is for nbin spoj or new binary spoj.

NBIN spoj New Binary spoj Here in the first we have to formulate the sequence as we can know how many bits can accommodate given “n”;
Now let formulate the table:-
Number of Bits
Binary number at given Nth position
Given N
1
1
1
2
10
2
3
100
3
3
101
4
4
1000
5
4
1001
6
4
1010
7
5
10000
8
5
10001
9
5
10010
10
5
10100
11
5
10101
12

From the above table we can see that five digit binary number can accommodate at most 5 values .
Now by carefully seeing this we can see that if remove the two Most significant Bit from five digit binary number we can see that rest binary number are the number less than three digit binary as see below


10
000
10
001
10
010
10
100
10
101
As we cliff the Left part of table .By the rest we can formulate the DP as:
Position of the array represent the number of Bits and content of array represent the maximum “N” that or less can be represented using that or less number of Bit :
PRE_COMPUTE[i] = PRE_COMPUTE[i-1] + PRE_COMPUTE[i-2] + 1;
Now for getting result take the string of MAX = 72(because (N = 1015  ) Nth number can be accommodated in at max 72 Bit you will get that after calculating) initially containing all ZEROES now for the given “N” find the minimum POSITION in array  which can accommodate Nth number and after that make
N = N - PRE_COMPUTE[POSITION -1] - 1 .And repeat till N>0;(you will get this if you use pen and paper and write few steps)

#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define LEN 72
LL pre_com[100];
void pre()
{
    pre_com[1] = 1;
    pre_com[2] = 2;
    pre_com[3] = 4;
    pre_com[4] = 7;
    int i=4;
    while(pre_com[i] <=1000000000000000LL)
    {
        i++;
        p[i] = p[i-2]+1;
        pre_com[i] =pre_com[i-1] + pre_com[i-2]+1;
    }
}
int find(LL n)
{
    int low=0,high=LEN,mid;
    mid = (low + high)>>1;
    while(true)
    {
        mid = (low + high)>>1;
        if(pre_com[mid] < n)
        {
            if(pre_com[mid + 1] >n)
                return mid;
            else
                low = mid;
        }
        else
        {
            if(pre_com[mid - 1] < n)
                return mid - 1;
            else
                high = mid;
        }
    }
}
void print(LL n)
{
    bool a[80]={false};
    int pos  = find(n);
    int max = pos + 1;
    a[pos+1]=true;
    LL diff = n - pre_com[pos] - 1;
    while(diff>0)
    {
        pos = find(diff);
        a[pos+1] = true;
        diff = diff - pre_com[pos] - 1;
    }
    for(int i=max ;i>=1;i--)
        if(a[i])
            printf("1");
        else
            printf("0");
}
int main()
{
    pre();
    int t;
    scanf("%d",&t);
    while(t--)
    {
        LL n;
        scanf("%lld",&n);
        print(n);
        printf("\n");
    }
    return 0;
}