D title
. 1 #include <the iostream> 2 #include <Map> . 3 #include <algorithm> . 4 #define int Long Long . 5 the using namespace STD; . 6 . 7 / * solving idea: by 10 fold in the title, only feint, you can not care, because as long as the size of the string on the line, a large number of strings of the same size multiplied by the natural or big 8 if there is only a maximum length of the input string string, such as input three characters string, namely: abc acb acbb naturally ACBB the maximum string, to find out the direct output 9 maximum string if a plurality of the same length in the input string, such as input four strings, respectively ab abcd abdc accc can be seen, abcd the maximum string, it is more the size of the string can be used directly 10 comparison operators to compare, but we want to single out the biggest, how to achieve it after three strings? 11 the maximum length string is inverted or less, using the reverse inverting function, the above example becomes: dcba cdba ccca again from the inside to find the maximum string, that is the final output is inverted again back dcba i.e. may be 12 * / 13 signed main() 14 { 15 int t; 16 cin >> t;//输入样例个数 17 while(t--) 18 { 19 map<int,int> vis;//使用一个图，相当于一个vis[]数组，用于存储最长字符串有多少个 20 string str[100];//开一个字符串数组，用于存储输入的字符串，本题只要大于10即可 21 string ans;//如果存在唯一一个最大字符串长度的字符串，则用于保存它，用于输出 22 int n; 23 int maxn = 0;//找出最长字符串 24 cin >> n;//输入字符串个数 25 for(int i = 1;i <= n;i++) 26 { 27 string temp; 28 cin >> temp;//循环输入n个字符串 29 int len = temp.size();//得到这个字符串的长度，进行下面比较操作 30 if(maxn < len) 31 { 32 ans = temp;//如果有出现暂时是最长的字符串，先记录到ans中 33 maxn = max(maxn,len);//更新maxn的值，使其保持最大状态 34 } 35 vis[len]++;//将此时的字符串长度记录到vis数组中，出现几个记录几个 36 reverse(temp.begin(),temp.end());//将字符串反转； 37 str[i] = temp;//将反转后的结果记录到str字符数组当中 38 } 39 //判断最长的字符串是否是一个，是，则直接输出，否则肯定存在多个相同长度的最长字符串 40 if(vis[maxn] == 1) 41 { 42 cout << ans << endl; 43 continue; 44 } 45 string res;//res相当于临时变量，用于存储最后答案 46 //以下的for操作是在反转后的字符串中通过字符串比较操作符直接选出最大的字符串 47 for(int i = 1;i <= n;i++) 48 { 49 if(str[i].size() == maxn)//找出最长的字符串 50 { 51 if(str[i] > res)//通过字符串比较操作符进行比较，选出大的记录在res当中 52 { 53 res = str[i]; 54 } 55 } 56 } 57 reverse(res.begin(),res.end());//由于得出的res是反转后的，必须再将其反转回来 58 cout << res << endl; 59 } 60 return 0; 61 }
E题
1 #include<iostream> 2 #define int long long 3 #define mod 2019 4 using namespace std; 5 int arr[10000000]; 6 7 signed main() 8 { 9 //在未开始输入的时候先对数据进行预处理，可大大降低时间复杂度 10 arr[0] = 3; 11 for(int i = 1;i <= 1000001;i++) 12 arr[i] = (arr[i - 1] * 2) % mod; 13 14 //等预处理完成后，再进行输入输出操作，就简便多了 15 int t; 16 cin >> t; 17 while(t--) 18 { 19 int n; 20 cin >> n; 21 cout << arr[n - 1] << endl; 22 } 23 return 0; 24 }