CF1063A Oh Those Palindromes

嘟嘟嘟

只要将每一种字母放一块输出就行了。

证明1：比如 1 2 3 4 5 6，那么这个序列对答案的贡献分别是1和5,2和4 ,3和6……如果重新排列成x x x x o o，会发现对          x x o x x o　 答案的贡献不变，所以得证。

证明2：字母ai有x个，那么对答案的最大贡献为x * (x - 1) / 2，重排后能达到理论上界，所以为最优解。　　　

 #include<cstdio>

 #include<iostream>

 #include<cmath>

 #include<algorithm>

 #include<cstring>

 #include<cstdlib>

 #include<cctype>

 #include<vector>

 #include<stack>

 #include<queue>

 using namespace std;

 #define enter puts("")

 #define space putchar(' ')

 #define Mem(a, x) memset(a, x, sizeof(a))

 #define rg register

 typedef long long ll;

 typedef double db;

 const int INF = 0x3f3f3f3f;

 const db eps = 1e-;

 const int maxn = 1e5 + ;

 inline ll read()

 {

   ll ans = ;

   char ch = getchar(), last = ' ';

   while(!isdigit(ch)) last = ch, ch = getchar();

   while(isdigit(ch)) ans = (ans << ) + (ans << ) + ch - '', ch = getchar();

   if(last == '-') ans = -ans;

   return ans;

 }

 inline void write(ll x)

 {

   if(x < ) x = -x, putchar('-');

   if(x >= ) write(x / );

   putchar(x %  + '');

 }

 int n, a[];

 char c[maxn];

 int main()

 {

   n = read(); scanf("%s", c);

   for(int i = ; i < n; ++i) a[c[i] - 'a']++;

   for(int i = ; i <= ; ++i)

     if(a[i]) while(a[i]--) putchar(i + 'a');

   enter;

   return ;

 }