C. Naming Company time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output Oleg the client and Igor the analyst are good friends. However, sometimes they argue over little things. Recently, they started a new company, but they are having trouble finding a name for the company. To settle this problem, they’ve decided to play a game. The company name will consist of n letters. Oleg and Igor each have a set of n letters (which might contain multiple copies of the same letter, the sets can be different). Initially, the company name is denoted by n question marks. Oleg and Igor takes turns to play the game, Oleg moves first. In each turn, a player can choose one of the letters c in his set and replace any of the question marks with c. Then, a copy of the letter c is removed from his set. The game ends when all the question marks has been replaced by some letter. For example, suppose Oleg has the set of letters {i, o, i} and Igor has the set of letters {i, m, o}. One possible game is as follows : Initially, the company name is ???. Oleg replaces the second question mark with ‘i’. The company name becomes ?i?. The set of letters Oleg have now is {i, o}. Igor replaces the third question mark with ‘o’. The company name becomes ?io. The set of letters Igor have now is {i, m}. Finally, Oleg replaces the first question mark with ‘o’. The company name becomes oio. The set of letters Oleg have now is {i}. In the end, the company name is oio. Oleg wants the company name to be as lexicographically small as possible while Igor wants the company name to be as lexicographically large as possible. What will be the company name if Oleg and Igor always play optimally? A string s = s1s2…sm is called lexicographically smaller than a string t = t1t2…tm (where s ≠ t) if si < ti where i is the smallest index such that si ≠ ti. (so sj = tj for all j < i) Input The first line of input contains a string s of length n (1 ≤ n ≤ 3·105). All characters of the string are lowercase English letters. This string denotes the set of letters Oleg has initially. The second line of input contains a string t of length n. All characters of the string are lowercase English letters. This string denotes the set of letters Igor has initially. Output The output should contain a string of n lowercase English letters, denoting the company name if Oleg and Igor plays optimally. Examples Input tinkoff zscoder Output fzfsirk Input xxxxxx xxxxxx Output xxxxxx Input ioi imo Output ioi Note One way to play optimally in the first sample is as follows : Initially, the company name is ???????. Oleg replaces the first question mark with ‘f’. The company name becomes f??????. Igor replaces the second question mark with ‘z’. The company name becomes fz?????. Oleg replaces the third question mark with ‘f’. The company name becomes fzf????. Igor replaces the fourth question mark with ‘s’. The company name becomes fzfs???. Oleg replaces the fifth question mark with ‘i’. The company name becomes fzfsi??. Igor replaces the sixth question mark with ‘r’. The company name becomes fzfsir?. Oleg replaces the seventh question mark with ‘k’. The company name becomes fzfsirk. For the second sample, no matter how they play, the company name will always be xxxxxx.
这道题有点儿坑，就是一个运用博弈论思想字符串排序的题。坑点在于当第一个字符串的最小值都比第二个字符串的最大值都大，因此需要将必须选的字符串（每个字符串选择一半即可）的最坏情况放在最后的位置。我用了6个伪指针来指向头尾位置，据说某大佬用vector轻轻松松就过了。。。还是自己太渣
#include<cstdio> #include<vector> #include<algorithm> #include<cstring> #include<iostream> using namespace std; char a[300005]; char b[300005]; char c[300005]; char cmp(char a, char b){ return a > b; } int main(){ cin>>a>>b; int n = strlen(a); sort(a, a + n); sort(b, b + n, cmp); int l = (n + 1) / 2; // l取长度的一半，即为每个字符串要选择的长度 int k = 0, p = n - 1; int x = l - 1, y = l - 1; int i = 0, j = 0; //k指向c字符串的头，p指向c字符串的尾 //i指向a字符串的头，x指向a字符串的尾 //j指向b字符串的头，y指向b字符串的尾 if(n%2 == 1) y--; // 坑点，如果n为奇数，b要选的长度减1 while(k <= p){ if(a[i] < b[j]) c[k++] = a[i++]; else //最坏情况 c[p--] = a[x--]; if(k > p) break; if(b[j] > a[i]) c[k++] = b[j++]; else //最坏情况 c[p--] = b[y--]; } for(int i = 0; i < n; i++) cout<<c[i]; cout<<endl; }