一、小易的升级之路

链接：小易的升级之路_牛客题霸_牛客网

简单题

#include<iostream>
using namespace std;
long long gcd(long long m, long long x)
{
	long long n ;
	while(x>0)
	{
		n = m % x;
		m = x;
		x = n;
	}
	return m;
}
int main()
{
	int n;
	long long m;
	cin >> n >> m;
	while (n--)
	{
		long long x;
		cin >> x;
		if (x <= m)
			m += x;
		else
			m += gcd(m, x);
	}
	cout << m;
}

二、礼物的最大价值

链接：礼物的最大价值_牛客题霸_牛客网

动态规划

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 grid int整型vector<vector<>>
     * @return int整型
     */

    int maxValue(vector<vector<int> >& grid) {
        int dp[202][202] = { 0 };
        for (int i = 1; i <= grid[0].size(); i++)
        {
            dp[1][i] = dp[1][i - 1] + grid[0][i - 1];
        }
        for (int i = 1; i <= grid.size(); i++)
        {
            dp[i][1] = dp[i - 1][1] + grid[i - 1][0];
        }
        for (int i = 2; i <= grid.size(); i++)
        {
            for (int j = 2; j <= grid[0].size(); j++)
            {
                dp[i][j] = max(dp[i][j - 1], dp[i - 1][j]) + grid[i - 1][j - 1];
            }
        }
        return dp[grid.size()][grid[0].size()];
    }
};

三、对称之美

直接头尾找相同字母，找不到说明没有

#include<iostream>
#include<string>
#include<stdio.h>
using namespace std;
int main()
{
	int n;
	cin >> n;
	while (n--)
	{
		int x;
		cin >> x;
		string s[102];
		for (int i = 0; i < x; i++)
		{
			string ss;
			cin >> ss;
			s[i] = ss;
		}
		int left = 0;
		int right = x - 1;
		while (left < right)
		{
			int flag = 0;
			for (int i = 0; i < s[left].size(); i++)
			{
				if (s[right].find(s[left][i]) != string::npos)
				{
					flag = 1;
					break;
				}

			}
			if (flag == 0)
				break;
			left++;
			right--;
		}
		if (left >= right)
			cout << "Yes" << endl;
		else
			cout << "No" << endl;

	}
}