LeetCode的第 363 场周赛——记录+补题

研究生生涯第一次打力扣周赛——3题

1. 计算 K 置位下标对应元素的和

class Solution {
public:
    int cnt(int x)
    {
        int sum = 0;
        while (x) {
            sum += ((x%2)?1:0);
            x/=2;
        }
        return sum;
    }
    int sumIndicesWithKSetBits(vector<int>& nums, int k) {
        int n = nums.size();
        int ans = 0;
        for (int i = 0;i < n;i ++) {
            if (cnt(i) == k) ans += nums[i];
        }
        return ans;
    }
};

2. 让所有学生保持开心的分组方法数

class Solution {
public:
    int countWays(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        int n = nums.size();
        nums.push_back(n+1);
        int ans = 0;
        if (nums[0] > 0) ans ++;
        for (int i = 0;i < n;i ++) {
            //if (nums[i] > i&&nums[i-1] < i) ans ++;
            if (nums[i] < (i+1)&&nums[i+1]>(i+1)) ans ++;
        }
        return ans;
    }
};

3. 最大合金数

二分模板

class Solution {
public:
    int maxNumberOfAlloys(int n, int k, int budget, vector<vector<int>>& composition, vector<int>& stock, vector<int>& cost) {
        long long ans = 0;
        for (int i = 0;i < k;i ++) {
            long long l = 0,r = 1000000000;
            while (l < r) {
                long long mid = (l+r+1)/2;
                long long cs = 0;
                for (int j = 0;j < n;j ++) {
                    if (mid*composition[i][j]>stock[j]) {
                        cs += (mid*composition[i][j]-stock[j])*cost[j];
                    } 
                }
                
                if (cs <= budget)
                    l = mid;
                else 
                    r = mid - 1;
            }
            ans = max(ans,l);
        }
        return ans;
    }
};

4. 完全子集的最大元素和

题意

思路

代码

#define MAXX ((int) 1e4)
bool inited = false;
bool flag[MAXX + 10];
int P[MAXX + 10];

// 全局预处理
void init() {
    if (inited) return;
    inited = true;

    memset(flag, 0, sizeof(flag));
    for (int i = 1; i <= MAXX; i++) P[i] = 1;
    // 筛法求质数
    for (int i = 2; i <= MAXX; i++) if (!flag[i]) {
        for (int j = i * 2; j <= MAXX; j += i) flag[j] = true;
        // 计算质数 i 在它的倍数 j 的质因数分解中出现了几次
        for (int j = i; j <= MAXX; j += i) {
            int tmp = j, cnt = 0;
            while (tmp % i == 0) tmp /= i, cnt++;
            if (cnt & 1) P[j] *= i;
        }
    }
}

class Solution {
public:
    long long maximumSum(vector<int>& nums) {
        init();
        int n = nums.size();
        // 用哈希表维护每组的和
        unordered_map<int, long long> mp;
        for (int i = 0; i < n; i++) mp[P[i + 1]] += nums[i];
        // 取每组和的最大值
        long long ans = 0;
        for (auto &p : mp) ans = max(ans, p.second);
        return ans;
    }
};

• 思路2
  
• 代码

class Solution {
public:
    long long maximumSum(vector<int>& nums) {
        long long retSum = 0, nowSum = 0;
        int n = nums.size();
        for (long long i = 1; i<=n; i++) {
            nowSum = nums[i-1];
            // cout<
            for (long long j = i+1; (j*j)/i <= n; j++) {
                if ((j*j)%i == 0 && (j*j)%(j*j/i) == 0) {
                    nowSum += nums[(j*j)/i-1];
                    // cout<
                }
            }
            // cout<
            if (nowSum > retSum) retSum = nowSum;
        }
        return retSum;
    }
};