A. Many Equal Substrings
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given a string t consisting of n lowercase Latin letters and an integer number k.
Let's define a substring of some string s with indices from l to r as s[l…r].
Your task is to construct such string s of minimum possible length that there are exactly k positions i such that s[i…i+n−1]=t. In other words, your task is to construct such string s of minimum possible length that there are exactly kk substrings of s equal to t.
It is guaranteed that the answer is always unique.
Input
The first line of the input contains two integers n and k (1≤n,k≤501≤n,k≤50) — the length of the string t and the number of substrings.
The second line of the input contains the string t consisting of exactly n lowercase Latin letters.
Output
Print such string s of minimum possible length that there are exactly k substrings of s equal to t.
It is guaranteed that the answer is always unique.
You are given a string t consisting of n lowercase Latin letters and an integer number k.
Let's define a substring of some string s with indices from l to r as s[l…r].
Your task is to construct such string s of minimum possible length that there are exactly k positions i such that s[i…i+n−1]=t. In other words, your task is to construct such string s of minimum possible length that there are exactly k substrings of s equal to t.
It is guaranteed that the answer is always unique.
Input
The first line of the input contains two integers n and k (1≤n,k≤50) — the length of the string t and the number of substrings.
The second line of the input contains the string t consisting of exactly n lowercase Latin letters.
Output
Print such string s of minimum possible length that there are exactly k substrings of s equal to t.
It is guaranteed that the answer is always unique.
Examples
input
3 4 aba
output
ababababa
input
3 2 cat
output
catcat
题意：给你一个字符串 t，和一个整数 k， 要你构造出一个最短的字符串 其中 t 出现的次数为 k
思路：这个题有很多细节值得注意， 首先是有循环节的情况 如 sprimsprimsprim 这样的，其次就是 aaaaa 这样的，然后有 abcab 这类的。如果使用 KMP求 出现次数 * 循环节长度 == 字符串 t 的长度 来判断 循环节的， 要注意还需要判断一次 ，如 xxxb, xx 的出现次数是2 ，它的长度也为2， 2 * 2 == 4, 但他不是循环节。
AC代码：
#include<bits/stdc++.h>
using namespace std;
int nex[100];
void getnext(char *t,int lent){
int i = 0,j = -1;
nex[i] = -1;
while(i < lent){
if(j == -1 || t[i] == t[j])
i ++,j ++, nex[i] = j;
else j = nex[j];
}
}
int kmp_count(char *t,char *s,int lent,int lens){
int i = 0,j = 0;
int ans = 0;
memset(nex,0,sizeof(nex));
getnext(t,lent);
while(i < lens){
if(j == -1 || s[i] == t[j])
i ++,j ++;
else j = nex[j];
if(j == lent){
i --; ans ++;
j = nex[j - 1];
}
}
return ans;
}
int main()
{
int n,k;
while(~scanf("%d%d",&n,&k)){
char t[100]; scanf("%s",t);
char tmp[100] = {0}; int ptm = 0;
int len = strlen(t), tim;
for(int i = 0;i < len;i ++){
tmp[ptm ++] = t[i];
tim = kmp_count(tmp,t,ptm,len);
if(tim * ptm == len) break;
}
char a[100] = {0};
for(int i = 0;i < tim;i ++) strcat(a,tmp);
int flag = 0;
if(kmp_count(tmp,a,ptm,ptm * tim) == tim && strcmp(tmp,t)) flag = 1;
if(flag){ ///有循环结
printf("%s",t);
for(int i = 0;i < k - 1;i ++) printf("%s",tmp);
}
else {
char f[100] = {0};
int p = len, wz = -1;
for(int i = len - 1;i >= 0;i --){ /// 寻找能够重复使用的字符，如abcab 中的 ab
f[-- p] = t[i];
int vis = 1, coun = 0;
for(int j = p;j < len;j ++)
if(f[j] != t[coun ++]){
vis = 0; break;
}
if(vis == 1 && p != 0) wz = p; /// p 等于0的话，意思就是寻到了本身，这里是针对单个字符的情况， 如样例 1 50 q .
}
if(wz == -1){
for(int i = 0;i < k;i ++) printf("%s",t);
}
else {
char xh[100] = {0}; int px = 0;
for(int i = len - wz;i < len;i ++) xh[px ++] = t[i]; ///如 abcab ，之前求得的是 ab, 但实际需要的是 cab, 所以是从 len - wz 开始
printf("%s",t);
for(int i = 0;i < k - 1;i ++) printf("%s",xh);
}
}
printf("\n");
}
return 0;
}