第 374 场 LeetCode 周赛题解

A 找出峰值

枚举

class Solution {
public:
    vector<int> findPeaks(vector<int> &mountain) {
        int n = mountain.size();
        vector<int> res;
        for (int i = 1; i < n - 1; i++)
            if (mountain[i] > mountain[i - 1] && mountain[i] > mountain[i + 1])
                res.push_back(i);
        return res;
    }
};

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B 需要添加的硬币的最小数量

贪心：对 $coins$ 升序排序，枚举 $coins[i]$ ，设当前已经可获得 $[0,reach]$ 范围内的任意整数金额，若 $coins[i]\le reach+1$ ，则由 $coins[0,i]$ 的子序列可得到 $[0,reach+coins[i]]$ 范围内的任意整数金额度，否则需要添加一个尽可能大的硬币 $reach+1$

class Solution {
public:
    int minimumAddedCoins(vector<int> &coins, int target) {
        sort(coins.begin(), coins.end());
        int res = 0;
        int e, i, reach;
        for (e = 1, i = 0, reach = 0; reach < target && i < coins.size();) {
            if (coins[i] <= reach + 1) {
                reach += coins[i++];
            } else {
                res++;
                reach += reach + 1;
            }
        }
        for (; reach < target;) {
            res++;
            reach += reach + 1;
        }
        return res;
    }
};

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C 统计完全子字符串

枚举 + 前缀和：首先用条件2将 $word$ 划分成若干满足子字符串，使得各子字符串内部任意相邻字符满足条件2，然后只需在各个子字符串中找满足条件1的子字符串，这样的字符串的长度只能是 $k,2k,\cdots ,26k$ 其中之一，枚举子字符串可能的长度 $len$，然后滑动窗口枚举场长为 $len$ 的子字符串，用前缀和判断在该子串中的字符数量是否为 $k$

class Solution {
public:
    int cmp(const string &s, int k) {
        int n = s.size();
        int pre[n + 1][26];//前缀和
        for (int j = 0; j < 26; j++)
            pre[0][j] = 0;
        for (int i = 0; i < n; i++)
            for (int j = 0; j < 26; j++)
                pre[i + 1][j] = pre[i][j] + (s[i] - 'a' == j ? 1 : 0);
        int res = 0;
        for (int len = k; len <= s.size() && len <= 26 * k; len += k) {
            for (int l = 0, r = len - 1; r < n; l++, r++) {//子串s[l,r]
                int can = 1;
                for (int i = 0; i < 26; i++) {
                    if (pre[r + 1][i] - pre[l][i] != 0 && pre[r + 1][i] - pre[l][i] != k) {
                        can = 0;
                        break;
                    }
                }
                if (can)
                    res++;
            }
        }
        return res;
    }

    int countCompleteSubstrings(string word, int k) {
        int res = 0;
        for (int i = 0, j = 0, n = word.size(); i < n;) {
            while (j + 1 < n && abs(word[j + 1] - word[j]) <= 2)
                j++;
            res += cmp(word.substr(i, j - i + 1), k);
            i = ++j;
        }
        return res;
    }
};

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D 统计感冒序列的数目

计数 + 逆元：设以感冒的人为间隔，未感冒的人可以划分成若干人数为 $a_i$ 的组：$a_1,\cdots ,a_k$，总序列数为：$\frac{(\sum a_i)!}{\prod a_i!}\times \prod f(a_i)$ ，其中当 $a_i$ 不是与边界相邻的组时 $f(a_i)=2^{a_i-1}$ ，否则 $f(a_i)=1$

class Solution {
public:
    using ll = long long;
    ll mod = 1e9 + 7;

    ll fpow(ll x, ll n) {//快速幂 x^n
        ll res = 1;
        for (ll e = x; n; e = e * e % mod, n >>= 1)
            if (n & 1)
                res = res * e % mod;
        return res;
    }

    ll inv(ll x) {//x 在 mod 下的匿元 
        return fpow(x, mod - 2);
    }

    int numberOfSequence(int n, vector<int> &sick) {
        vector<ll> fact(n + 1, 1), f2(n + 1, 1);
        for (int i = 1; i <= n; i++) {
            fact[i] = fact[i - 1] * i % mod;
            f2[i] = f2[i - 1] * 2 % mod;
        }
        int res = 1;
        int m = sick.size();
        int s = 0;
        if (sick[0] != 0) {//与左边界相邻的组
            s += sick[0];
            res = res * inv(fact[sick[0]]) % mod;
        }
        if (sick[m - 1] != n - 1) {//与右边界相邻的组
            s += n - 1 - sick[m - 1];
            res = res * inv(fact[n - 1 - sick[m - 1]]) % mod;
        }
        for (int i = 0; i < m - 1; i++)//枚举中间的各个分组
            if (sick[i + 1] != sick[i] + 1) {
                int t = sick[i + 1] - sick[i] - 1;
                s += t;
                res = res * inv(fact[t]) % mod;
                res = res * f2[t - 1] % mod;
            }
        res = res * fact[s] % mod;
        return (res + mod) % mod;
    }
};

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