T1 100/100

组合数学，先判断哪些小火龙可以打到怪物，然后用总方案减去每一条小火龙都没打到怪物的情况即可

typedef long long ll;
class Solution {
public:
    int methodsOfKillingMonster(vector<vector<int> >& a, vector<int>& b) {
        ll mod=1e9+7;
        auto qpow=[&](ll x,ll y) {
            ll ans=1;
            while(y) {
                if(y%2==1) ans=ans*x%mod;
                x=x*x%mod;
                y/=2;
            }
            return ans;
        };
        int cnt=0;
        for(auto i:a) if(i[0]==b[0]||i[1]==b[1]) ++cnt;
        ll ans=qpow(4,a.size());
        ans=(ans-qpow(3,cnt)*qpow(4,a.size()-cnt)%mod+mod)%mod;
        return ans;
    }
};

T2 100/100

这题是真打表找规律

typedef long long ll;
class Solution {
public:
    int countOfE(int n) {
        ll mod=1e9+7;
        auto qpow=[&](ll x,ll y) {
            ll ans=1;
            while(y) {
                if(y%2==1) ans=ans*x%mod;
                x=x*x%mod;
                y/=2;
            }
            return ans;
        };
        ll ans=qpow(25,n-1)*n%mod;
        return ans;
    }
};

T3 44/100

没怎么理解构造方案，事后根别人讨论说也许可以先处理出左右子树的个数，再根据个数填入

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
public:
    TreeNode* dfs(int l,int r) {
        if(l>r) return nullptr;
        int len=r-l+1;
        TreeNode* now=new TreeNode(-1);
        if(len%2==0) {
            now->val=l+len/2;
        }
        else {
            if(len==1) now->val=l;
            else if(len==3) now->val=l+1;
            else now->val=l+len/2+1;
        }
        now->left=dfs(l,now->val-1);
        now->right=dfs(now->val+1,r);
        return now;
    }
    
    TreeNode* maxDepthAVL(int n) {
        return dfs(1,n);
    }
};

T4 100/100

st表维护区间最小值+单调栈找下一个大于当前数字的数

typedef long long ll;
class Solution {
public:
    vector<int> getEmotion(vector<int>& a) {
        int n=a.size();
        vector<vector<ll>> st(n,vector<ll>(32,0x3f3f3f3f));
        for(int i=0;i<n;++i) st[i][0]=a[i];
        for(int j=1;j<32;++j)
            for(int i=0;i+(1ll<<j)-1<n;++i)
                st[i][j]=min(st[i][j-1],st[i+(1ll<<(j-1))][j-1]);
        auto lg=[&](int x) {
            ll ans=0;
            ll tmp=1;
            while(tmp*2<x) {
                tmp*=2;
                ++ans;
            }
            return ans;
        };
        auto mn=[&](int l,int r) {
            int len=r-l+1;
            int j=lg(len);
            return min(st[l][j],st[r-(1ll<<j)+1][j]);
        };
        vector<int> flg(n);
        stack<int> s;
        for(int i=n-1;i>=0;--i) {
            while(!s.empty()&&a[i]>=a[s.top()]) s.pop();
            if(s.empty()) flg[i]=n-1;
            else flg[i]=s.top();
            s.push(i);
        }
//         for(int i=0;i<n;++i) cout<<flg[i]<<' ';cout<<endl;
        vector<int> ans(n);
        for(int i=0;i<n;++i) ans[i]=a[i]-mn(i,flg[i]);
        return ans;
    }
};