2166. Design bit set

2166. Design bit set

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subject

BitSet Bitset It is a data structure that can store bits in a compact form .
Please realize Bitset class .

Bitset(int size) use size Bit initialization Bitset , All the seats are 0 .
void fix(int idx) Mark the subscript as idx The value on the bit of is updated to 1 . If the value is already 1 , Nothing will change .
void unfix(int idx) Mark the subscript as idx The value on the bit of is updated to 0 . If the value is already 0 , Nothing will change .
void flip() Flip Bitset The value on each bit in . let me put it another way , All values are 0 The bit will become 1 , vice versa .
boolean all() Check Bitset in every Whether the values of are all 1 . If this condition is met , return true ; otherwise , return false .
boolean one() Check Bitset in whether At least one The value of is 1 . If this condition is met , return true ; otherwise , return false .
int count() return Bitset The median is 1 A bit of total .
String toString() return Bitset The current composition of . Be careful , In the result string , The first i Characters at subscripts should be the same as Bitset No i It's the same .

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Code

class Bitset {
    
public:
    int n;   //  Total length 
    int sum; // 1 The number of 
    bool fp; //  reverse 
    vector<bool> st;

    Bitset(int size) {
    
        st.resize(size);
        n = size;
        fp = false;
        sum = 0;
    }
    
    void fix(int idx) {
    
        // idx Set up 1
        // fp==false: st[idx]==false, sum++
        // fp==true: st[idx]==true, sum++
        if(!(st[idx] ^ fp)) sum++, st[idx] = !st[idx];
    }
    
    void unfix(int idx) {
    
        // idx Set up 0
        if(st[idx] ^ fp) sum--, st[idx] = !st[idx];
    }
    
    void flip() {
    
        fp = !fp;
        sum = n - sum;
    }
    
    bool all() {
    
        return sum == n;
    }
    
    bool one() {
    
        return sum >= 1;
    }
    
    int count() {
    
        return sum;
    }
    
    string toString() {
    
        string res = "";
        for (int i = 0; i < n; i++) 
            if(st[i] ^ fp) res += '1';
            else res += '0';
        return res;
    }
};