C. Cow and Message（思维）

outputstandard output
Bessie the cow has just intercepted a text that Farmer John sent to Burger Queen! However, Bessie is sure that there is a secret message hidden inside.

The text is a string s of lowercase Latin letters. She considers a string t as hidden in string s if t exists as a subsequence of s whose indices form an arithmetic progression. For example, the string aab is hidden in string aaabb because it occurs at indices 1, 3, and 5, which form an arithmetic progression with a common difference of 2. Bessie thinks that any hidden string that occurs the most times is the secret message. Two occurrences of a subsequence of S are distinct if the sets of indices are different. Help her find the number of occurrences of the secret message!

For example, in the string aaabb, a is hidden 3 times, b is hidden 2 times, ab is hidden 6 times, aa is hidden 3 times, bb is hidden 1 time, aab is hidden 2 times, aaa is hidden 1 time, abb is hidden 1 time, aaab is hidden 1 time, aabb is hidden 1 time, and aaabb is hidden 1 time. The number of occurrences of the secret message is 6.

Input
The first line contains a string s of lowercase Latin letters (1≤|s|≤105) — the text that Bessie intercepted.

Output
Output a single integer — the number of occurrences of the secret message.

Examples
inputCopy
aaabb
outputCopy
6
inputCopy
usaco
outputCopy
1
inputCopy
lol
outputCopy
2
Note
In the first example, these are all the hidden strings and their indice sets:

a occurs at (1), (2), (3)
b occurs at (4), (5)
ab occurs at (1,4), (1,5), (2,4), (2,5), (3,4), (3,5)
aa occurs at (1,2), (1,3), (2,3)
bb occurs at (4,5)
aab occurs at (1,3,5), (2,3,4)
aaa occurs at (1,2,3)
abb occurs at (3,4,5)
aaab occurs at (1,2,3,4)
aabb occurs at (2,3,4,5)
aaabb occurs at (1,2,3,4,5)
Note that all the sets of indices are arithmetic progressions.
In the second example, no hidden string occurs more than once.

In the third example, the hidden string is the letter l.
思路：其实看样例就可以看出来，字符串长度越长，数据越小，所以我们只需要计算长度为1和2的时候。写循环的时候墨迹了好久转不过弯来。注意用long long
代码如下：

#include
#define ll long long
using namespace std;

const int maxx=30;
vector<int> p[maxx];
string s;
int n;

int main()
{
	cin>>s;
	n=s.length();
	for(int i=0;i<=26;i++) p[i].clear();
	for(int i=0;i<n;i++) p[s[i]-'a'].push_back(i);
	ll _max=0;
	for(int i=0;i<26;i++) _max=max(_max,(ll)p[i].size());
	for(int i=0;i<26;i++) _max=max(_max,(ll)(p[i].size()*((ll)p[i].size()-1)/2));
	for(int j=0;j<26;j++)
	{
		for(int i=0;i<26;i++)
		{
			if(i==j) continue;
			ll zz=0;
			for(int k=0;k<p[i].size();k++)
			{
				int pos=lower_bound(p[j].begin(),p[j].end(),p[i][k])-p[j].begin();
				zz+=((ll)p[j].size()-(ll)pos);
			}
			_max=max(_max,zz);
		}
	}
	cout<<_max<<endl;
	return 0;
}//aababaabaaaa

努力加油a啊，(^o)/~