一、压缩字符串

链接：压缩字符串(一)_牛客题霸_牛客网

#include<iostream>
using namespace std;
#include<vector>
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param param string字符串
     * @return string字符串
     */
    string compressString(string param) {

        string s;
        s = param;
        vector<char> a;
        vector<int> num;
        for (int i = 0; i < s.size(); i++)
        {
            if ((s[i] <= 'z' && s[i] >= 'a') || (s[i] >= 'A' && s[i] <= 'Z'))
            {
                if (a.empty() || s[i] != a.back())
                {
                    a.push_back(s[i]);
                    num.push_back(1);
                }
                else
                    num.back()++;
            }
        }
        string s2;
        for (int i = 0; i < a.size(); i++)
        {
            s2.push_back((a[i]));
            if (num[i] != 1)
            {
                int k = num[i];
                string s4;
                while (k)
                {
                    s4.push_back(k % 10 + '0');
                    k /= 10;
                }
                string::reverse_iterator it = s4.rbegin();
                while (it != s4.rend())
                {
                    s2.push_back(*it);
                    it++;
                }
            }
        }
        return s2;
    }
};

二、chika和蜜柑

先用甜度排序，再酸度排序，取前K个

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
#include<map>
struct mcomp
{
	bool operator()(pair<long long, long long>& a, pair<long long, long long>& b)
	{

		if (a.first == b.first)
			return a.second > b.second;
		else
			return false;
	}
}mcomp;
int main()
{
	int n, k;
	cin >> n >> k;
	vector<long long> so;
	vector<long long> sw;
	for (int i = 0; i < n; i++)
	{
		long long x;
		cin >> x;
		so.push_back(x);
	}
	for (int i = 0; i < n; i++)
	{
		long long x;
		cin >> x;
		sw.push_back(x);
	}
	multimap<long long, long long, greater<long long>> m;

	for (int i = 0; i < n; i++)
	{
		m.insert(pair<long long, long long>(sw[i], so[i]));
	}
	vector < pair < long long, long long >> v(m.begin(), m.end());
	sort(v.begin(), v.begin() + k, mcomp);

	long long max1 = 0;
	long long max2 = 0;
	for (int i = 0; i < k; i++)
	{
		max1 += v[i].first;
		max2 += v[i].second;
	}
	cout << max2 << ' ' << max1;
}

三、01背包问题

链接：01背包_牛客题霸_牛客网

老问题了，直接复制粘贴

class Solution
{
 int dp[1010] = { 0 };
public:
 int knapsack(int V, int n, vector<vector<int> >& vw) 
 {
 for(int i = 0; i < n; i++)
 {
 for(int j = V; j >= vw[i][0]; j--)
 {
 dp[j] = max(dp[j], dp[j - vw[i][0]] + vw[i][1]);
 }
 }
 return dp[V];
 }
};