第 91 场双周赛

1 不同的平均值数目

不同的平均值数目

Solution

class Solution {
public:
    int distinctAverages(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        int i = 0,j = nums.size() - 1;
        int vis[205];
        memset(vis,0,sizeof(vis));
        int ans = 0;
        for(;i < j;i++, j--){
            int t = nums[i] + nums[j];
            if(!vis[t]) ans++;
            vis[t] = 1;
        }
        return ans;
    }
};

2 统计构造好字符串的方案数

统计构造好字符串的方案数

Solution

class Solution {
public:
    int countGoodStrings(int low, int high, int zero, int one) {
        long long mod = 1e9+7;
        long long ans = 0;
        long long dp[100005] = {0};
        dp[0] = 1;
        for(int i = 0;i < high + 1;i++){
            if(i >= zero) dp[i] = (dp[i] + dp[i - zero]) % mod;
            if(i >= one) dp[i] = (dp[i] + dp[i - one]) % mod;
            if(i >= low) ans = (ans + dp[i]) % mod;
        }
        return ans;
    }
};

3 树上最大得分和路径

树上最大得分和路径

Solution

class Solution {
public:
    int mostProfitablePath(vector<vector<int>>& edges, int bob, vector<int>& amount) {
        //建图
        int n = amount.size();
        vector<int> e[n];
        for(auto &edge: edges){
            e[edge[0]].push_back(edge[1]);
            e[edge[1]].push_back(edge[0]);
        }
        // bfs开始跑
        int pa[n],A[n],B[n];//分别表示 节点i的祖先 A到节点i需要几步 B到节点i需要几步
        function<void(int, int, int)> dfs1 = [&](int sn, int fa, int d){
            pa[sn] = fa;
            A[sn] = d;
            for (int fn : e[sn]) if (fn != fa) dfs1(fn, sn, d + 1);
        };
        dfs1(0, -1, 0);

        //从bob不断向上走，计算B[i]
        for(int i = 0;i < n;i++) B[i] = n + 1;
        for(int sn = bob,i = 0;sn >= 0;sn = pa[sn], i++) B[sn] = i;
        //
        int ans = -2e9;
        function<void(int, int)> dfs2 = [&](int sn, int tot){
            if (A[sn] < B[sn]) tot += amount[sn];
            else if (A[sn] == B[sn]) tot += amount[sn] / 2;
            int ch = 0;
            for (int fn : e[sn]) if (fn != pa[sn]){
                ch++;ch++;
                dfs2(fn, tot);
            }
            if (ch == 0) ans = max(ans, tot);
        };
        dfs2(0, 0);
        return ans;
    }
};

4 根据限制分割消息

根据限制分割消息

Soltion

class Solution {
public:
    vector<string> splitMessage(string message, int limit) {
        int n = message.length();
        for (int i = 1, cap = 0, tail_len;; ++i) {
            if (i < 10) tail_len = 5; // 结尾的长度
            else if (i < 100) {
                if (i == 10) cap -= 9; // 前面的结尾的长度都 +1，那么容量就要减小
                tail_len = 7;
            } else if (i < 1000) {
                if (i == 100) cap -= 99;
                tail_len = 9;
            } else {
                if (i == 1000) cap -= 999;
                tail_len = 11;
            }
            if (tail_len >= limit) return {}; // cap 无法增大，寄
            cap += limit - tail_len;
            if (cap < n) continue; // 容量没有达到，继续枚举

            vector<string> ans(i);
            for (int j = 0, k = 0; j < i; ++j) {
                string tail = "<" + to_string(j + 1) + "/" + to_string(i) + ">";
                if (j == i - 1) ans[j] = message.substr(k) + tail;
                else {
                    int m = limit - tail.length();
                    ans[j] = message.substr(k, m) + tail;
                    k += m;
                }
            }
            return ans;
        }
    }
};