Problem:
This is the hard version of the problem. The only difference is the constraints on n and k. You can make hacks only if all versions of the problem are solved.
You have a string s, and you can do two types of operations on it:
Delete the last character of the string.
Duplicate the string: s:=s+s
, where +
denotes concatenation.
You can use each operation any number of times (possibly none).
Your task is to find the lexicographically smallest string of length exactly k that can be obtained by doing these operations on string s.
A string ais lexicographically smaller than a string b
if and only if one of the following holds: a is a prefix of b, but a≠b;
In the first position where a
and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
The first line contains two integers n, k (1≤n,k≤5⋅105) — the length of the original string sand the length of the desired string.
The second line contains the string s, consisting of n lowercase English letters.
Output
Print the lexicographically smallest string of length k
that can be obtained by doing the operations on string s.
Example
Input
8 16
dbcadabc
Output
dbcadabcdbcadabc
Input
4 5
abcd
Output
aaaaa
思路：
先从第二个字母开始遍历，与第一个字母比较，若大则直接标记上然后break，若相等则以两个相同字母开始，向后比较，直到出现不等为止，然后分类处理，详见code。
code:
#include <bits/stdc++.h>
#define DEBUG freopen("_in.txt", "r", stdin);
// #define DEBUG freopen("_in.txt", "r", stdin),freopen("_out.txt", "w", stdout);
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const ll maxn = 5e5 + 10;
const ll maxm = maxn * maxn;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll mod = 2147493647;
ll t, n, m, k;
char str[maxn], str1[maxn];
int main()
{
// DEBUG;
scanf("%lld%lld", &n, &k);
scanf("%s", str + 1);
ll pos=0;
ll i=2;
while (i<=n)
{
if (str[i] > str[1])
{
pos = i-1;
break;
}
else if (str[i] == str[1])
{
ll j = 1, num = i;
while (i<=n&& j < num&&str[i] == str[j] )
{
i++;//这里i比j小了，就一定比第一个小，保证O(n)
j++;
}
if(i>n)
{
pos = num-1;
break;
}
if (j == num)
continue;
if(str[i]<str[j]) continue;
pos = num-1;
break;
}
else
{
i++;
}
}
if(pos==0)
{
pos=n;
}
for (ll i = 1; i <= k; i++)
{
if (i % pos != 0)
printf("%c", str[i % pos]);
else
printf("%c", str[pos]);
}
printf("\n");
return 0;
}