题意：

给你一个字符串，问其子串中有多少个满足：
1.子串头尾字母不相同；
2.子串内部字母与头尾字母不相同；
3.子串长度大于等于2；
问有多少个这样的字串？

题目：

Leo has started a job in a travel agency. His first task is to organize a summer trip to an exotic overseas city. During the summer season, events of various types take place in the city: sports matches, concerts, beach parties, and many others. At any given time, there is exactly one event taking place. Events of any particular type may take place more than once during the season. The itinerary of events that Leo offers to his clients cannot be chosen arbitrarily; the company requires them to form a so-called “good itinerary.” A good itinerary is a consecutive sequence of at least two events in the summer season, where the first and last events are of different types, and they are both unique among all event types during the sequence. For example, if the first event in a good itinerary is a beach party, none of the other events during the itinerary can also be a beach party. There are no other restrictions on the event types in the sequence of a good itinerary.

Before he starts organizing the trip, Leo wants to know the total number of good itineraries that are possible given a calendar of events that will take place over the summer season.

输入描述:

The input consists of one line with a string describing the sequence of event types in the summer season. All characters are lowercase English letters (a - z)(a−z), with different letters represent different types of events. Character ii of the string encodes the ii-th event of the summer. There are no blanks or spaces in the string.

The length of the input string is at least 22 and at most 100 000100000 characters.

输出描述:

Print the number of good itineraries that exist for the given summer season.
示例1

输入

abbcccddddeeeee

输出

10
示例2

输入

thenumberofgoodstringsis

输出

143

分析：

由题意我们很容易可以得出：
1.连续重复字母是没有意义的，只有第一个字母起作用，其他重复因题目条件字串内部不能和子串首位相同，满足题意只有两种情况:1.连续重复的某字母在子串内；2.只有一个字母为首或尾，所以可以用set去重，找不重复字母；
2.当之前遍历的字母中存在重复，
例如：thenumbe
那么对于最后一个e来说，求得是字符串numbe的子串满足题意；
3.类单调栈思维，将一个字母，入所有字母的栈中，当遍历到某字母时，加上set去重后的有效前缀的长度即可，维护有效前缀的方法是，由于上一次字母及其前缀都是不可用的，当遍历到某字母加和后，直接清空栈即可。

AC代码：

#include<stdio.h>
#include<string.h>
#include<stack>
#include<set>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn=1e5+10;
char a[maxn];
int main()
{int l;int ans=0;scanf("%s",a);set<int >s[30];l=strlen(a);for(int i=0;i<l;++i){int x=a[i]-'a'+1;ans+=s[x].size();for(int j=1;j<=26;++j)s[j].insert(x);s[x].clear();}printf("%d\n",ans);return 0;
}