SPOJ:String Play (？)

String Play

 

Milo has a string S of length L. Tutu picks a random prefix and Mota picks a random suffix ofS.

Now, Chotku is given a task of concatenating the two strings that Tutu and Mota have chosen, in respective order. Chotku wonders, how many distinct prefix-suffix concatenation is possible out there of string S.

So you know what to do.. Help Chotku!

Input

Input file contains several lines of text. Each line contains a string, S. End of file marks the end of input.

Output

Output one integer per string, denoting the number of distinct prefix-suffix concatenation of the string.

Constraints

1. Strings consist of lower-case letters only. 

2. 1 ≤ L ≤ 10000006

 

 

Sample Input	Output for Sample Input
abc aab	8 7

 

Explanation:

For sample #1, the 3 prefixes are “a”, “ab”, “abc”

The 3 suffixes are “c”, “bc”, “abc”

And the 8 distinct concatenations are, “ac”, “abc”, “aabc”, “abbc”, “ababc”, “abcc”, “abcbc”, “abcabc”.

 

题意：一个字符串，A取一段前缀，B取一段后缀，然后连接起来组成一个新的字符串S，问可以组成多少种新的字符串，重复的只统计一次。

思路：首先必须是O(n)的复杂度，然后我什么都不知道了。

 

#include<bits/stdc++.h>
const int maxn=10000010;
using namespace std;
char c[maxn];  int num[27];
int main()
{
    int Len,i; long long ans;
    while(~scanf("%s",c+1)){
        Len=strlen(c+1); ans=Len;
        memset(num,0,sizeof(num));
        for(i=1;i<=Len;i++) num[c[i]-'a']++; 
        for(i=Len-1;i>=1;i--){
            ans+=Len-num[c[i+1]-'a'];
            if(c[i+1]==c[Len]) ans++;
        }
        printf("%lld\n",ans);
    }
    return 0;
}