第81场双周赛

统计星号

简单模拟

class Solution {
    
public:
    int countAsterisks(string s) {
    
        int cnt=0;
        vector<int>res;
        for(int i=0;i<s.size();i++)
        {
    
            if(s[i]=='*') cnt++;
            else if(s[i]=='|') res.push_back(i);
        }

        for(int i=0;i<res.size();i+=2)
        {
    
            int a=res[i],b=res[i+1];
            for(int j=a;j<=b;j++)
            {
    
                if(s[j]=='*') cnt--;
            }
        }
        return cnt;
    }
};

统计无向图中无法互相到达点对数

并查集

class Solution {
    
typedef long long LL;
public:
    vector<int>p,s;
    int find(int x)
    {
    
        if(p[x]!=x) p[x]=find(p[x]);
        return p[x];
    }
    long long countPairs(int n, vector<vector<int>>& edges) {
    
        for(int i=0;i<n;i++)
        {
    
            p.push_back(i);
            s.push_back(1);
        }

        for(auto&x:edges)
        {
    
            int a=find(x[0]),b=find(x[1]);
            if(a!=b)
            {
    
                s[b]+=s[a];
                p[a]=b;
            }
        }

        LL res=n*(n-1ll)/2;
        for(int i=0;i<n;i++)
        {
    
            if(p[i]==i)
            {
    
                res-=s[i]*(s[i]-1ll)/2;
            }
        }
        return res;
    }
};

操作后的最大异或和

class Solution {
    
public:
    int maximumXOR(vector<int>& nums) {
    
        int ans=0;
        for(auto &x:nums)
        {
    
            ans|=x;
        }
        return ans;
    }
};

不同骰子序列的数目

dp y总的答案，我不会我好菜。。。。

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 (u != j && u != k && gcd(u + 1, j + 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;
    }
};