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_i个，那么对答案的最大贡献为x_i * (x_i - 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 ;
 }