字母出现次数的奇偶关系，可以用状态压缩和位异或的形式表示。

由于数据范围是0~2^25，没法直接开数组，所以用到map容器

需要注意的是：如果仅仅只是为了查看map<key,value>中键是否存在，最好使用map::find(key)函数，直接用map[key]的值来判断会格外增加一倍的时间

#include <cstdio>
#include <map>
using namespace std;
const int MAXN = 20005;
int T, n, res, pos, z, k, o, p;
int AA[30];
char ss[MAXN];
map<int,int> mp1, mp2;
map<int,int>::iterator it;
void solve(map<int,int> &mp, map<int,int> &mq)
{
	int i, j;
	for (it = mp.begin(); it!=mp.end();++it)
	{
		z = it->second; k = it->first;
		for (i = 0; i<26; ++i)
		{
			j = k^AA[i];
			if (mq.find(j) != mq.end())
				res += z*mq[j];
		}
		res += z*mq[k];
	}
}
int main()   
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
#endif
	for (int i = 0; i< 26; ++i) AA[i] = 1<<i;
	scanf("%d", &T);	
	while (T--)
	{		
		scanf("%s", ss);
		mp1.clear(); mp2.clear();
		pos = -1;
		for ( n = 0; ss[n]; ++n)
		if (ss[n] == '?')
			pos = n;
		res = pos!=-1;
		p = 0;
		for (int i = pos-1; i>= 0; --i)
		{
			o = ss[i]-'a';
			p = p^AA[o];
			if (!p) ++res;
			res += mp1[p];
			mp1[p]++;
		}
		p = 0;
		for (int i = pos+1; i< n; ++i)
		{
			o = ss[i]-'a';
			p = p^AA[o];
			if (!p) ++res;
			res += mp2[p];
			mp2[p]++;
		}
		if (pos != -1)
		{
			for (it = mp1.begin(); it!=mp1.end();++it)
			{
				z = it->second; k = it->first;
				if (0 == (k & (k-1))) res += z;
			}
			for (it = mp2.begin(); it!=mp2.end();++it)
			{
				z = it->second; k = it->first;
				if (0 == (k & (k-1))) res += z;
			}
			if (mp1.size() < mp2.size()) solve(mp1, mp2);
			else solve(mp2, mp1);
		}
		printf("%d\n", res);
	}
    return 0;
}