Cyclic NacklaceTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4063 Accepted Submission(s): 1830 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 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
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 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 Author possessor WC Source HDU 3rd “Vegetable-Birds Cup” Programming Recommend lcy | We have carefully selected several similar problems for you: 1358 2222 3068 2896 3744 |
题意：给你一个字符串 头尾可以连接（这句没什么乱用） 问你最少添加几个字符，会组成两个循环的字符串
比如abca 在最后加bc 就是一个两个循环的字符串
解题思路：
这题主要考察KMP中next数组的使用~next数组就是用来存最长公共前缀的，所以我们只要求出了最长公共前缀L 用串的长度l%L就得到剩下的 然后用L 再减去剩下的 就得到最少要加的珠子。
AC代码：
#include <stdio.h>
#include <string.h>
#include <algorithm>
#define maxn 100000+10
using namespace std;
char P[maxn];
int pre[maxn];
void getnext(int n)
{
int i=0,j=-1;
pre[0]=-1;
while(i<n){
if(j==-1||P[i]==P[j]){
pre[++i]=++j;
}
else j=pre[j];
}
}
int main()
{
int t;
scanf("%d",&t);
while(t--){
scanf("%s",P);
int l=strlen(P);
getnext(l);
int L=l-pre[l];//循环节长度
if(l%L==0&&L!=l){
printf("0\n");
continue;
}
int res=L-l%L;
printf("%d\n",res);
}
return 0;
}
Cyclic NacklaceTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4063 Accepted Submission(s): 1830 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 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
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 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 Author possessor WC Source HDU 3rd “Vegetable-Birds Cup” Programming Recommend lcy | We have carefully selected several similar problems for you: 1358 2222 3068 2896 3744 |
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