STL-bitset

Introduction to bitset-personal

I can't find a accurate explanation online.So there is my understanding of it.
Bitset is a STL which is easy to treat a number as a binary.

Instructions

It is included by include<bitset>
If you want to define a bitset,the code is as follows

bitset<num> n;//num must be defined as an integer literal constant or an integer type const object initialized with a constant value.

If so initialized, each bit is 0.
Of course you can also define it like this

int a;
bitset<num> n(a);

now,the bitset n is the binary form of a.And you even can use this like a array.
Such as

#include <iostream>
#include <bitset>
using namespace std;
int main()
{
    int a;
    cin>>a;
    bitset<4> b(a);
    cout<<b<<endl;
    cout<<b[0]<<endl<<b[1]<<endl<<b[2]<<endl<<b[3]<<endl;
    return 0;
}

Input:
13 -5
Ouput:
1101
1
0
1
1

1011
1
1
0
1

As can be seen from these examples, when the bit position in the bitset is operated, it is reversed.
Such as:

int n = 16;
bitset<64> b(n);
b[0]=0
b[1]=0
b[2]=0
b[3]=0
b[4]=1

From the above example we can see that the negative number will directly become the complement form under the action of bitset.This has advantages and disadvantages.This has advantages and disadvantages, because some places just let you represent the binary form, no need to complement.If you meet a situtation like this,just take a judge,all numbers are processed as positive numbers, if they are negative, then add a negative sign.I'd like to paste a example here.

Example

Source: 简单的二进制
Description：
计算机只能识别0和1,使用的是二进制，而在日常生活中人们使用的是十进制，”正如亚里士多德早就指出的那样，今天十进制的广泛采用，只不过是我们绝大多数人生来具有10个手指头这个解剖学事实的结果。

在计算机领域中，补码是一个重要的概念，数值一律用补码来表示和存储。原因在于，使用补码，可以将符号位和数值域统一处理；同时，加法和减法也可以统一处理。此外，补码与原码相互转换，其运算过程是相同的，不需要额外的硬件电路。

学习了计算机导论后，相信大家都明白了原码、反码和补码的区别，并能快速地将一个十进制整数的补码写出。今天我们将问题进行简化：给出一个十进制的整数N，请写出N在64位二进制补码表示下，一共有多少个1。

Input
多组输入，每一行有一个整数N（保证N在int64范围内）

Output
输出N在二进制补码下1的个数，每组输出占一行

Examples
input
10
15
output
2
4
Note
正整数的补码是其二进制表示，与原码相同

求负整数的补码，将其原码除符号位外的所有位取反后加一

Problem solving：
We can see this problem need the complement when the number is negative.So the bitset is so comfortable for this.

Code:

#include<bits/stdc++.h>
using namespace std;
int main()
{
    long long n;
    while(scanf("%lld",&n)!=EOF)
    {
        bitset<64> b(n);
        cout<<b.count()<<endl;
    }
    return 0;
}

If you write this, the code will be very simple.The specific function functions will be discussed below.

Many functions

Almost all STL have lots of functions,bitset is no exception.
I don't have the ability, and I don't have the energy to list them all, so I will list what I like and what I find useful.

Overview

Define a bitset first:
bitset<64> b;

Next I will help you understand the usage of these functions with some examples.

b.any() and b.none() and b.count()

b.any(): Is there a binary bit with a value of 1 in b?

#include<iostream>
#include<bitset>
using namespace std;
int main()
{
    int n=5,m=0;// n = 0101, m = 0
    bitset<8> b1(n);
    bitset<8> b2(m);
    cout<<b1.none()<<" "<<b2.none();
}
Operation result:
1 0

So you can see that the return value of b.any() is the bool type.

b.none()：Is there no binary bit with a value of 1 in b?

#include<iostream>
#include<bitset>
using namespace std;
int main()
{
    int n=5,m=0;// n = 0101, m = 0
    bitset<8> b1(n);
    bitset<8> b2(m);
    cout<<endl<<b1.none()<<" "<<b2.none();
}
Operation result:
0 1

So you can see that the return value of b.none() also is the bool type.

b.count(): The number of bits in b with a value of 1

#include<iostream>
#include<bitset>
using namespace std;
int main()
{
    int n=15,m=16;// n = 1111, m = 10000
    bitset<8> b1(n);
    bitset<8> b2(m);
    cout<<b1.count()<<" "<<b2.count();
}
Operation result:
4 1

So the function of b.count() is to return the number of 1 in the binary.Just like the example above

b[pos]

b[pos]: Access the binary bit at pos in b
I started the function's introduction at the beginning of this blog, you can look up.

b.test(pos)

b.test(pos): Whether the binary bit in pos is 1 in b？

#include<iostream>
#include<bitset>
using namespace std;
int main()
{
    int n=15,m=16;// n = 1111, m = 10000
    bitset<8> b1(n);
    bitset<8> b2(m);
    cout<<b1.test(1)<<" "<<b2.test(4)<<" "<<b2.test(1);
}
Operation result:
1 1 0

So you can see that the return value of b.test(pos) is the bool type,too.

b.set() and b.set(pos)

b.set(): Set all binary bits in b to 1

#include<iostream>
#include<bitset>
using namespace std;
int main()
{
    int n = 16;
    bitset<8> b(n);
    cout<<b<<endl;
    b.set();
    cout<<b;
}
Operation result:
00010000
11111111

b.set(pos): Put the binary position in b at pos as 1

#include<iostream>
#include<bitset>
using namespace std;
int main()
{
    int n = 16;
    bitset<8> b(n);
    cout<<b<<endl;
    b.set(0);
    cout<<b;
}
Operation result:
00010000
00010001

b.reset() and b.reset(pos)

b.reset(): Set all binary bits in b to 0
b.reset(pos): Put the binary position in b at pos to 0
These two functions are very similar to the two above, so I won’t go into details.

b.flip() and b.flip(pos)

b.flip(): All binary bits in b are inverted bit by bit

#include<iostream>
#include<bitset>
using namespace std;
int main()
{
    int n = 16;
    bitset<8> b(n);
    cout<<b<<endl;
    b.flip();
    cout<<b;
}
Operation result:
00010000
11101111

b.flip(pos): Invert the binary bit at b in pos
Basically the same as above, no longer repeat

b.to_string()

b.to_string(): Convert the contents of the bitset to string

#include<iostream>
#include<bitset>
using namespace std;
int main()
{
    int n = 16;
    bitset<8> b(n);
    string s;
    s=b.to_string();
    cout<<s<<endl;
}
Operation result:
00010000

b.to_ulong()

#include<iostream>
#include<bitset>
using namespace std;
int main()
{
    int n = 16;
    bitset<5> b(n);
    b.flip();
    long a;
    cout<<b<<endl;
    a=b.to_ulong();
    cout<<a<<endl;
}
Operation result:
01111
15