2020 LCP 力扣团队赛

1. 黑白方格画

二进制枚举

class Solution {
public:
    int paintingPlan(int n, int k) {
        map<int,int>ans;
        if(k == n*n)return 1;
        
        for(int i = 0;i < (1<<n);i++){
            for(int j = 0;j < (1<<n);j++){
                int x = __builtin_popcount(i);
                int y = __builtin_popcount(j);
 
                ans[(x+y)*n-x*y]++;
            }
        }
        return ans[k];
    }
};

2. 魔术排列

纯模拟

class Solution:

    def isMagic(self, target: List[int]) -> bool:
        n = len(target)  
        x = [i+1 for i in range(n)]
        
        idx = 0
        cnt = 0
        k = -1
        while idx < n:
            x = x[cnt:][1::2] + x[cnt:][::2]
            cnt = 0
            
            if x[cnt] != target[idx]:
                return False
            c = 0
            while idx < n and x[cnt] == target[idx]:
                cnt += 1
                idx += 1
                c += 1
            if k == -1:
                k = c
            else:
                if c != k:
                    
                    if idx == n:
                        return True
                    return False
            
        return True
                
        

3. 数字游戏

两个堆维护中位数

class Solution {
public:
    const int mod = 1e9+7;
    vector<int> numsGame(vector<int>& nums) {
        priority_queue<int,vector<int>,greater<int>>y;
        priority_queue<int,vector<int>,less<int>>x;
        for(int i = 0;i < nums.size();i++){
            nums[i] -= i;
        }
        vector<int>ans;
        long long sum1 = 0,sum2 = 0;
        for(int i : nums){
            if(x.empty())x.push(i),sum1+=i;
            else{
                int tp = x.top();
                if(i <= tp){
                    x.push(i);
                    sum1 += i;
                }
                else{
                    y.push(i);
                    sum2 += i;
                }
                while(x.size() > y.size() + 1){
                    y.push(x.top());sum2 += x.top();
                    sum1 -= x.top();x.pop();
                }
                if(x.size() == y.size()-1){
                    x.push(y.top());sum1 += y.top();
                    sum2 -= y.top();y.pop();
                }
            }
            long long mid = x.top();
        
            //cout << (sum2-y.size()*mid) + (x.size()*mid - sum1) % mod << endl;
            ans.push_back(((sum2-y.size()*mid) + (x.size()*mid - sum1)) % mod);
        }
        return ans;
        
    }
};

4. 古董键盘

别人的一份代码

class Solution {
private:
    /* 计算组合数 */
    int C(int m, int n) {
        if ((n == 0) || (n == m))
            return 1;
        
        int ret = 1;
        
        for (int i = 1; i <= n; ++i) {
            ret *= (m - i + 1);
            ret /= i;
        }

        return ret;
    }

public:
    int keyboard(int k, int n) {
        vector<long long> dp(n + 1, 0);

        /* 按键0次则得出唯一的一种内容: 空 */
        dp[0] = 1;

        /* 遍历26个字母(26种货物) */
        for (int i = 0; i < 26; ++i) {
            /* 遍历背包的容量(从大到小) */
            for (int j = n; j >= 1; --j) {
                /* 放入x个当前字母 */
                for (int x = 1; x <= k; ++x) {
                    if (x > j)
                        continue;
                    dp[j] += C(j, x) * dp[j - x];
                    dp[j] %= 1000000007;
                }
            }
        }

        return dp.back();
    }
};