Problem Description
CC always becomes very depressed at the end of this month, he has checked his credit card yesterday, without any surprise, there are only 99.9 yuan left. he is too distressed and thinking about how to tide over the last days. Being inspired by the entrepreneurial spirit of “HDU CakeMan”, he wants to sell some little things to make money. Of course, this is not an easy task.
As Christmas is around the corner, Boys are busy in choosing christmas presents to send to their girlfriends. It is believed that chain bracelet is a good choice. However, Things are not always so simple, as is known to everyone, girl’s fond of the colorful decoration to make bracelet appears vivid and lively, meanwhile they want to display their mature side as college students. after CC understands the girls demands, he intends to sell the chain bracelet called CharmBracelet. The CharmBracelet is made up with colorful pearls to show girls’ lively, and the most important thing is that it must be connected by a cyclic chain which means the color of pearls are cyclic connected from the left to right. And the cyclic count must be more than one. If you connect the leftmost pearl and the rightmost pearl of such chain, you can make a CharmBracelet. Just like the pictrue below, this CharmBracelet’s cycle is 9 and its cyclic count is 2:
Now CC has brought in some ordinary bracelet chains, he wants to buy minimum number of pearls to make CharmBracelets so that he can save more money. but when remaking the bracelet, he can only add color pearls to the left end and right end of the chain, that is to say, adding to the middle is forbidden.
CC is satisfied with his ideas and ask you for help.
Input
The first line of the input is a single integer T ( 0 < T <= 100 ) which means the number of test cases.
Each test case contains only one line describe the original ordinary chain to be remade. Each character in the string stands for one pearl and there are 26 kinds of pearls being described by ‘a’ ~‘z’ characters. The length of the string Len: ( 3 <= Len <= 100000 ).
Output
For each case, you are required to output the minimum count of pearls added to make a CharmBracelet.
Sample Input
3
aaa
abca
abcde
Sample Output
0
2
5
思路
題目說了一堆的廢話，大概意思就是給定一個字符串，問至少加入多少個字符可以使得這個字符串變成一個由k個重複子串組成的字符串(k > 1)。樣例一不需要加入因爲可以由3個a組成輸出0，樣例二加入b和c字符即可滿足要求輸出2，樣例三加入abcde一組字符串輸出5。
針對KMP算法的next數組做一個小小的加強深刻理解next數組的含義，個人認爲KMP算法的真正用途不在於單模匹配(Sunday和BM做單模匹配性能更優)，而next數組纔是精髓。當匹配失敗時回到next[j]的位置，也就是說將j的下標位向右移動了j - next[j]位，也就是說這j - next[j]個字符是一組重複的字符串。利用這一性能可以找到一個字符串的最小循環節長度(j - next[j])同理也可以找到循環節個數。
- 最小循環節長度是len - next[len];
- 完整的循環節的個數是 len / (len - next[len])
- 剩餘不夠一個循環節補全需要的字符個數 (len - next[len]) - len % (len - next[len])
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
const int maxn = 1e5+1e2;
char s[maxn];
int p[maxn];
void get_next(int len)
{
memset(p,-1,sizeof(p));
int i = -1,j = 0;
while(j < len){ //最好是多算一位，多算一位不影響做匹配但是做Next數組拓展好用
if(i == -1 || s[i] == s[j]){
i++;j++;
p[j] = i;
}
else{
i = p[i];
}
}
}
int main()
{
int n;
cin>>n;
while(n--){
scanf("%s",s);
int len = strlen(s);
get_next(len);
int res = len - p[len];
if(len % res == 0 && p[len] != 0){ //循環節個數至少是兩個。
cout<<0<<endl;
}
else{
cout<<res - (len%res)<<endl;
}
}
return 0;
}
願你走出半生，歸來仍是少年~