# HDU 3746 Cyclic Nacklace

## 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

### 题目链接

求使输入字符串成为其某个子串(循环节)循环连接的字符串所最少添加的字符。

此题使用到KMP算法中的next数组，KMP算法详解:

### 从头到尾彻底理解KMP

其中next数组为字符串“前缀”和“后缀”的最长共有元素的长度，next数组详解:

### 字符串匹配的KMP算法

设字符串长度为len，那么最小循环节即为len-next[len]，详解:

### kmp next函数 kmp的周期问题，深入了解kmp中next的原理

## AC代码:

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/*-----------------------------------------------------------*/ /* Blog :henuly.top */ /*-----------------------------------------------------------*/ #include <bits/stdc++.h> using namespace std; #define mem(a,b) memset(a,b,sizeof(a)) #define pb push_back #define mp make_pair #define lowbit(x) (x&(-x)) #define XDebug(x) cout<<#x<<"="<<x<<endl; #define ArrayDebug(x,i) cout<<#x<<"["<<i<<"]="<<x[i]<<endl; #define print(x) out(x);putchar('\n') typedef long long ll; typedef unsigned long long ull; typedef pair<int,int> PII; typedef pair<double,double> PDD; typedef pair<ll,ll> PLL; const int INF = 0x3f3f3f3f; const int maxn = 1e5 + 5; const int mod = 1e9 + 7; const double eps = 1e-8; const double pi = asin(1.0) * 2; const double e = 2.718281828459; template <class T> inline bool read(T &ret) { char c; int sgn; if (c = getchar(), c == EOF) { return 0; } while (c != '-' && (c < '0' || c > '9')) { c = getchar(); } sgn = (c == '-') ? -1 : 1; ret = (c == '-') ? 0 : (c - '0'); while (c = getchar(), c >= '0' && c <= '9') { ret = ret * 10 + (c - '0'); } ret *= sgn; return 1; } template <class T> inline void out(T x) { if (x < 0) { putchar('-'); x = -x; } if (x > 9) { out(x / 10); } putchar(x % 10 + '0'); } // 求next数组 void GetNext(char *str, vector<int> &Next) { int i = 0, j = -1; Next[0] = -1; int len = strlen(str); while (i != len) { if (j == -1 || str[i] == str[j]) { Next[++i] = ++j; } else { j = Next[j]; } } } int main(int argc, char *argv[]) { #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); freopen("out.txt", "w", stdout); #endif int T; read(T); for (int Case = 1; Case <= T; ++Case) { char str[maxn]; scanf("%s", str); int len = strlen(str); vector<int> Next(len + 1, 0); GetNext(str, Next); // 循环节长度 int CycleLength = len - Next[len]; if (len != CycleLength && len % CycleLength == 0) { print(0); } else { print(CycleLength - Next[len] % CycleLength); } } #ifndef ONLINE_JUDGE fclose(stdin); fclose(stdout); system("gedit out.txt"); #endif return 0; } |

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