腾讯笔试消消乐与字符串AC代码

第一题消消乐，每次把出现次数最多的两个次数相减，只要保证最后优先队列没有数，就是YES
第四题字符串，求最小循环节，只要循环节一样，就能匹配，cnt++，不一样则无法循环
剩下的几道都只A了一点，有AC的大佬分享下思路吗~

# define  _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cmath>
#include <queue>
#include <map>
#include <algorithm>
#include <stack>
#include <iomanip>
#include <vector>
#include <unordered_map>
#include <string>
#include <set>
using namespace std;

typedef long long LL;
#define fo(i, a, b) for (unsigned int i = a; i < b; ++i)
#define fb(i, a, b) for (unsigned int i = a; i > b; --i)
#define min(a, b) ((a) < (b)? (a) : (b))
#define max(a, b) ((a) > (b)? (a) : (b))
const LL MOD = 1000000000007;

void tencentOne()
{
    int t; cin >> t;
    for (int i = 0; i < t; ++i)
    {
        int n; cin >> n;
        unordered_map<int, int> um;
        int a;
        for (int j = 0; j < n; ++j)
        {
            cin >> a;
            if (um.find(a) == um.end()) um[a] = 1;
            else um[a]++;
        }
        priority_queue<int, vector<int>, less<int>> pq;
        for (auto& u : um)
            pq.emplace(u.second);
        while (pq.size() > 1)
        {
            int a = pq.top();
            pq.pop();
            int b = pq.top();
            pq.pop();
            a -= b;
            if (a) pq.emplace(a);
        }
        cout << (pq.size() ? "NO\n" : "YES\n");
    }
}


void getNext(string& s, int len, vector<int>& kmp)
{
    int i = 0;
    int j = -1;
    kmp[0] = -1;
    while (i < len)
    {
        if (-1 == j || s[i] == s[j]) kmp[++i] = ++j;
        else j = kmp[j];
    }
}

void tencentFour()
{
    for (int n; cin >> n;)
    {
        string s; cin >> s;
        int m; cin >> m;
        int len = s.length();
        vector<int> kmp(len + 1, 0);
        getNext(s, len, kmp);
        int period = len - kmp[len];
        if (len % (len - kmp[len])) period = len;
        string periodstr = s.substr(0, period);
        int cnt = 0;
        fo(i, 0, m)
        {
            cin >> s;
            int len = s.length();
            kmp.resize(len + 1);
            getNext(s, len, kmp);
            int p = len - kmp[len];
            if (len % (len - kmp[len])) p = len;
            if (s.substr(0, p) == periodstr) cnt++;
        }
        cout << cnt << endl;
    }
}


int main()
{
    tencentOne();
    tencentFour();
    return 0;
}