Cyclic Nacklace
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2514 Accepted Submission(s): 1136
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 colZ??http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">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"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
KMP的简单应用：提供几组样例： abbabbabbab abcabcabc
#include
#include
#include
#include
#include
#include
#include
using namespace std; const int maxn = 100000+10; int next[maxn],n,ans; char str[maxn]; void getNext(){ next[0] = 0,next[1] = 0; for(int i = 1; i < n; i++){ int j = next[i]; while(j && str[i] != str[j]) j = next[j]; if(str[i]==str[j]) next[i+1] = j+1; else next[i+1] = 0; } } int main(){ int ncase; cin >> ncase; while(ncase--){ scanf("%s",str); n = strlen(str); getNext(); if(n%(n-next[n])==0&&next[n]!=0){ ans = 0; }else{ if(next[n]==0) ans = n; else{ int t = n-next[n]; ans = t - next[n]%t; } } printf("%d\n",ans); } return 0; }
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