Let‘s call a string adorable if its letters can be realigned in such a way that they form two consequent groups of equal symbols (note that different groups must contain different symbols). For example, ababa is adorable (you can transform it to aaabb, where the first three letters form a group of a

You‘re given a string s. Check whether it can be split into two non-empty subsequences such that the strings formed by these subsequences are adorable. Here a subsequence is an arbitrary set of indexes of the string.

The only line contains s (1 ≤ |s| ≤ 105) consisting of lowercase latin letters.

Print «Yes» if the string can be split according to the criteria above or «No» otherwise.

Each letter can be printed in arbitrary case.

ababa

Yes

zzcxx

Yes

yeee

No

題意：如果能把一個字符串分成兩份，每份都可以被分成只有相同字母的兩塊，並且這兩塊的字母不相同，那麽輸出yes，反之no。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define lowbit(x) (x&(-x))
#define eps 0.00000001
#define pn printf("\n")
using namespace std;
typedef long long ll;

const int maxn = 1e5+7;
map <char,int> mp;

int main()
{
    char s[maxn];
    gets(s);
    int len = strlen(s);
    for(int i=0;i<len;i++)
        mp[s[i]]++;
    int fl = 0;
    for(map<char,int>::iterator it = mp.begin(); it!=mp.end(); it++)
        if(it->second == 1)
            fl++;
    if(mp.size() > 4) printf("No\n");
    else if(mp.size() == 2 && fl) printf("No\n");
    else if(mp.size() == 1) printf("No\n");
    else if(mp.size() == fl && fl%2 == 1) printf("No\n");
    else printf("Yes\n");
    
}

Codeforces Round #471 (Div. 2)B. Not simply beatiful strings