Li Kou's 81st biweekly match

6104. Statistical asterisk

class Solution {
    
public:
    int countAsterisks(string s) {
    
        int cnt=1;
        int res=0;
        for(auto x:s)
        {
    
            if(x=='|')cnt++;
            else
            {
    
                if(cnt%2)
                {
    
                    if(x=='*')res++;
                }
            }
        }    
        return res;

    }
};

6106. Count the logarithm of points that cannot reach each other in an undirected graph

class Solution {
    
public:
    int f[100010];
    int cnt[100010];
    int find(int x)
    {
    
        if(x!=f[x])f[x]=find(f[x]);
        return f[x];
    }
    void unite(int a,int b)
    {
    
        int x=find(a),y=find(b);
        if(x!=y)
        {
    
            cnt[x]+=cnt[y];
            f[y]=x;
        }
    }
    long long countPairs(int n, vector<vector<int>>& edges) {
    
        
        for(int i=0;i<n;i++)f[i]=i,cnt[i]=1;
        for(int i=0;i<edges.size();i++)
        {
    
            int a=edges[i][0],b=edges[i][1];
            unite(a,b);
        }
        vector<int>v;
        for(int i=0;i<n;i++)
            if(f[i]==i)v.push_back(i);
        long long res=0;
        long long sum=0;
        for(int i=0;i<v.size();i++)
        {
    
            res+=(long long)cnt[v[i]]*sum;
            sum+=cnt[v[i]];
        }
        return res;
    }
};

6105. Maximum XOR and after operation

class Solution {
    
public:
    int maximumXOR(vector<int>& nums) {
    
        int cnt[32]={
    0};
        for(int i=0;i<nums.size();i++)
        {
    
            int x=nums[i];
            for(int j=0;j<31;j++)
            {
    
                if((x>>j)&1)cnt[j]++;
            }
        }
        for(int i=0;i<31;i++)
            if(cnt[i])cnt[i]=1;
        int s=1;
        int sum=0;
        for(int i=0;i<31;i++)
        {
    
            if(cnt[i])sum+=s*cnt[i];
            s<<=1;

        }
        return sum;
    }
};

6107. The number of different die sequences

const int N=10010,mod=1e9+7;
int f[N][6][6];
class Solution {
    
public:
    int gcd(int a,int b)
    {
    
        return b?gcd(b,a%b):a;
    }
    int distinctSequences(int n) {
    
        memset(f,0,sizeof f); 
        if(n==1)return 6;
        for(int i=0;i<6;i++)
            for(int j=0;j<6;j++)
                if(gcd(i+1,j+1)==1&&i!=j)
                    f[2][i][j]=1;
        for(int i=3;i<=n;i++)
            for(int j=0;j<6;j++)
                for(int k=0;k<6;k++)
                    if(j!=k&&gcd(j+1,k+1)==1)
                        for(int u=0;u<6;u++)
                            if(k!=u&&u!=j&&gcd(j+1,u+1)==1)
                                f[i][j][k]=(f[i][j][k]+f[i-1][u][j])%mod;
        int res=0;
        for(int i=0;i<6;i++)
            for(int j=0;j<6;j++)
                res=(res+f[n][i][j])%mod;
        return res;
    }
};