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力扣第 393 场周赛

A

讨论即可

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class Solution {
public:
    string findLatestTime(string s) {
        if (s[1] == '?')
            s[1] = s[0] == '0' ? '9' : '1';
        if (s[0] == '?')
            s[0] = s[1] >= '2' ? '0' : '1';
        if (s[3] == '?')  s[3] = '5';
        if (s[4] == '?')  s[4] = '9';
        return s;
    }
};

B

两边找

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class Solution {
public:
    int maximumPrimeDifference(vector<int>& nums) {
        int n=nums.size(),a,b;
        for(int i=0;i<n;++i){
            if(nums[i]!=1&&sushu(nums[i])){
                a=i;
                break;
            }
        }
        for(int i=n-1;i>-1;--i){
            if(nums[i]!=1&&sushu(nums[i])){
                b=i;
                break;
            }
        }
        return b-a;
    }
    
    bool sushu(int m){
        int i,k;
        k=(int)sqrt( (double)m );
        for(i=2;i<=k;i++){
            if(m%i==0)
                break;
        }
        if(i>k) return true;
        return false;
    }
};


C

枚举子集，然后容斥，奇加偶减

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class Solution {
public:
    long long findKthSmallest(vector<int>& coins, int k) {
        auto check = [&](long long m) -> bool {
            long long cnt = 0;
            for (int i = 1; i < (1 << coins.size()); i++) { // 枚举所有非空子集
                long long lcm_res = 1; // 计算子集 LCM
                for (int j = 0; j < coins.size(); j++) {
                    if (i >> j & 1) {
                        lcm_res = lcm(lcm_res, coins[j]);
                        if (lcm_res > m) { // 太大了
                            break;
                        }
                    }
                }
                cnt += __builtin_popcount(i) % 2 ? m / lcm_res : -m / lcm_res;
            }
            return cnt >= k;
        };

        long long left = k - 1, right = (long long) ranges::min(coins) * k;
        while (left + 1 < right) {
            long long mid = (left + right) / 2;
            (check(mid) ? right : left) = mid;
        }
        return right;
    }
};


D

这题我也没懂太彻底

整体是st表维护dp

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const int N = 10001, L = 18, M = 11;
const int INF = 1000000000;

class Solution {
private:
    int v[L][N], log[N], mi[L][N];
    int f[N];

    int get_v(int l, int r) {
        int z = log[r - l];

        return v[z][l] & v[z][r - (1 << z) + 1];
    }

    int binary_search(int ed, int target) {
        int l = 1, r = ed + 1;

        while (l < r) {
            int mid = (l + r) >> 1;
            if (get_v(mid, ed) < target)
                l = mid + 1;
            else
                r = mid;
        }

        return l;
    }

    int get_min(int l, int r) {
        int z = log[r - l];

        return min(mi[z][l], mi[z][r - (1 << z) + 1]);
    }

public:
    int minimumValueSum(vector<int>& nums, vector<int>& andValues) {
        const int n = nums.size(), m = andValues.size();

        log[0] = 0;
        for (int i = 1; i <= n; i++) {
            v[0][i] = nums[i - 1];
            log[i] = (i >> (log[i - 1] + 1)) ? log[i - 1] + 1 : log[i - 1];
        }

        for (int j = 1; (1 << j) <= n; j++)
            for (int i = 1; i + (1 << j) - 1 <= n; i++)
                v[j][i] = v[j - 1][i] & v[j - 1][i + (1 << (j - 1))];

        for (int i = 1; i <= n; i++)
            mi[0][i] = f[i] = INF;

        mi[0][0] = f[0] = 0;

        for (int k = 1; (1 << k) <= n; k++)
            for (int i = 0; i + (1 << k) - 1 <= n; i++)
                mi[k][i] = min(mi[k - 1][i], mi[k - 1][i + (1 << (k - 1))]);

        for (int j = 0; j < m; j++) {
            f[0] = INF;
            for (int i = 1; i <= n; i++) {
                int p1 = binary_search(i, andValues[j]);
                int p2 = binary_search(i, andValues[j] + 1);

                if (p1 < p2)
                    f[i] = get_min(p1 - 1, p2 - 2) + nums[i - 1];
                else
                    f[i] = INF;
            }

            for (int i = 0; i <= n; i++) {
                mi[0][i] = f[i];
                f[i] = INF;
            }

            for (int k = 1; (1 << k) <= n; k++)
                for (int i = 0; i + (1 << k) - 1 <= n; i++)
                    mi[k][i] = min(mi[k - 1][i], mi[k - 1][i + (1 << (k - 1))]);
        }

        if (mi[0][n] >= INF)
            return -1;

        return mi[0][n];
    }
};