# 2's complement

### Problem Statement :

```Understanding 's complement representation is fundamental to learning about Computer Science. It allows us to write negative numbers in binary. The leftmost digit is used as a sign bit. If it is , we have a negative number and it is represented as the two's complement of its absolute value. Let's say you wrote down the 's complement representation for each -bit integer in the inclusive range from  to . How many 's would you write down in all?

For example, using an -bit byte rather than  bit integer, the two's complement of a number can be found by reversing all its bits and adding . The two's complement representations for a few numbers are shown below:

To write down that range of numbers' two's complements in  bits, we wrote 's. Remember to use  bits rather than  in your solution. The logic is the same, so the  bit representation was chosen to reduce apparent complexity in the example.

Function Description

Complete the twosCompliment function in the editor below. It should return an integer.

twosCompliment has the following parameter(s):
- a: an integer, the range minimum
- b: an integer, the range maximum

Input Format

The first line contains an integer , the number of test cases.

Each of the next  lines contains two space-separated integers,  and .

Output Format

For each test case, print the number of 's in the -bit 's complement representation for integers in the inclusive range from  to  on a new line.```

### Solution :

```                            ```Solution in C :

In  C  :

#include<stdio.h>
long long num[34];
long long ONES[33];
void init()
{
int i;
ONES[0]=1;
ONES[1]=2;
num[0]=1;
for(i=1;i<33;i++)
num[i]=num[i-1]<<1;
for(i=2;i<32;i++)
ONES[i]=(ONES[i-1]<<1)+(num[i]-num[i-1])-1;
}

long long calc(int x)
{
long long sum=0,temp=0;
int i,j,k;
if(x==0)
return 0;
for(i=0;i<33;++i)
{
if(x<num[i])
break;
}
i--;
sum=ONES[i];
temp=x-num[i];
return sum+calc(temp)+temp;
}

void solve(long long a,long long b)
{
long long lo,hi,count;
if(a<0)
{
a++;
lo=calc(-a);
lo=(32*(-a))-lo;
lo+=32;
a--;
}
else
{
if(!a)
lo=calc(a);
else
lo=calc(a-1);
}
if(b<0)
{
if(b==-2)
hi=32;
else if(b==-1)
hi=0;
else
{
b+=2;
hi=calc(-b);
hi=(32*(-b))-hi;
hi+=32;
b-=2;
}
}
else
hi=calc(b);
if(a<0&&b>0)
count=hi+lo;
else
count=hi-lo;
if(count<0)count=-count;
printf("%lld\n",count);
}
int main()
{
int t,a,b;
init();
scanf("%d",&t);
while(t--)
{
scanf("%d %d",&a,&b);
solve(a,b);
}
return 0;
}```
```

```                        ```Solution in C++ :

In  C ++ :

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std ;
#define INF (int)1e9

long long solve(int a)
{
if(a == 0) return 0 ;
if(a % 2 == 0) return solve(a - 1) + __builtin_popcount(a) ;
return ((long long)a + 1) / 2 + 2 * solve(a / 2) ;
}

long long solve(int a,int b)
{
if(a >= 0)
{
long long ret = solve(b) ;
if(a > 0) ret -= solve(a - 1) ;
return ret ;
}
long long ret = (32LL * -(long long)a) - solve(~a) ;
if(b > 0) ret += solve(b) ;
else if(b < -1)
{
b++ ;
ret -= (32LL * -(long long)b) - solve(~b) ;
}
return ret ;
}

int main()
{
int runs,a,b ;
cin >> runs ;
while(runs--)
{
cin >> a >> b ;
long long ret = solve(a,b) ;
cout << ret << endl ;
}
return 0 ;
}```
```

```                        ```Solution in Java :

In  Java :

import java.io.IOException;
import java.util.HashMap;
import java.util.Map;

public class Solution {

private static Map<Long, Long> onesCache = new HashMap<Long, Long>();

private static long computeOnes(long n, long div) {
long onesCount = 0;

if (onesCache.containsKey(n))
return onesCache.get(n);
else if (n == 0)
return 0;
else if (n == 1)
return 1;

long a = n / div;
long b = n % div;

if (a == 1) {
onesCount += b + 1;
onesCount += computeOnes(div - 1, div >> 1);
}
onesCount += computeOnes(b, div >> 1);

onesCache.put(n, onesCount);

return onesCount;
}

private static long computeOnes(long n) {
if (n < 0)
return -n * 32 - computeOnes(-n - 1, 1<<30);
else
return computeOnes(n, 1<<30);
}

/**
* @param args
* @throws IOException
*/
public static void main(String[] args) throws IOException {
String line;

try {
int tests = Integer.parseInt(line);

int[][] intervals = new int[tests][2];
for (int i=0; i < tests; i++) {
String[] tokens = line.split(" ");
intervals[i][0] = Integer.parseInt(tokens[0]);
intervals[i][1] = Integer.parseInt(tokens[1]);
}

for (int i = 0; i < intervals.length; i++) {
int x = intervals[i][0];
int y = intervals[i][1];

if ((x > 0 && y > 0) || (x < 0 && y < 0)) {
long n1 = computeOnes(x + (x > 0 ? -1 : 0));
long n2 = computeOnes(y + (y < 0 ? +1 : 0));
System.out.println(Math.abs(n1 - n2));
} else {
long n1 = computeOnes(x);
long n2 = computeOnes(y);
System.out.println(n1 + n2);
}
}
} catch (IOException e) {
e.printStackTrace();
} finally {
br.close();
}
}

}```
```

```                        ```Solution in Python :

In  Python3 :

t = int(input())

def gen(num):
if num == '':
return 0
elif num == '1':
return 1
elif num == '0':
return 0
elif num[0] == '0':
return(gen(num[1:]))
else:
return(int(num[1:],2)+1+gen(num[1:])+2**(len(num)-2)*(len(num)-1))

def func(a,b):
if a < 0 and b >= 0:
if a+b+1 == 0: return(32*(b+1))
elif a+b+1 < 0: return(32*(b+1)+func(a,-(b+2)))
else: return(32*(-a)+func(-a,b))
elif a < 0 and b < 0:
return(32*(b-a+1)-func(-b-1,-a-1))
else:
if a == 0:
return gen(bin(b)[2:])
return gen(bin(b)[2:]) - gen(bin(a-1)[2:])

for ts in range(t):
a,b = input().strip().split();a,b = int(a),int(b)
print(func(a,b))```
```

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