Foj 2075 Substring

Calculate the string with the smallest Lexicographic Order that exactly appears n times.

Question: When the suffix array is added with a monotonous stack and n is set to 1, it must be judged. However, the data is a bit watery and can be passed without being judged.

[Cpp]
# Include <stdio. h>
# Include <stdlib. h>
# Include <string. h>
# Include <string>
# Include <queue>
# Include <algorithm>
# Include <vector>
# Include <stack>
# Include <list>
# Include <iostream>
# Include <map>
Using namespace std;
# Define inf 0x3f3f3f
# Define M 100010
Int max (int a, int B)
{
Return a> B? A: B;
}
Int min (int a, int B)
{
Return a <B? A: B;
}
Int height [M], rank [M], r [M], sa [M];
Int ts [M], ta [M], tb [M], TV [M], pos, st, ed;
Struct node
{
Int h, st, w;
} Q [M];
Bool cmp (int * y, int a, int B, int l)
{
Return y [a] = y [B] & y [a + l] = y [B + l];
}
Void da (int n, int m)
{
Int I, j, * x = ta, * y = tb, p;
For (I = 0; I <m; I ++) ts [I] = 0;
For (I = 0; I <n; I ++) ts [x [I] = r [I] ++;
For (I = 1; I <m; I ++) ts [I] + = ts [I-1];
For (I = n-1; I> = 0; I --) sa [-- ts [x [I] = I;
 
For (j = 1, p = 1; p <n; j * = 2, m = p)
{
P = 0;
For (I = n-j; I <n; I ++) y [p ++] = I;
For (I = 0; I <n; I ++) if (sa [I]> = j) y [p ++] = sa [I]-j;
For (I = 0; I <m; I ++) ts [I] = 0;
For (I = 0; I <n; I ++) TV [I] = x [y [I];
For (I = 0; I <n; I ++) ts [TV [I] ++;
For (I = 1; I <m; I ++) ts [I] + = ts [I-1];
For (I = n-1; I> = 0; I --) sa [-- ts [TV [I] = y [I];
Swap (x, y );
X [sa [0] = 0;
P = 1;
For (I = 1; I <n; I ++)
{
If (cmp (y, sa [I-1], sa [I], j) x [sa [I] = p-1;
Else x [sa [I] = p ++;
}
}
}
Void calh (int n)
{
Int I, k, tmp;
For (I = 1; I <= n; I ++) rank [sa [I] = I;
K = 0;
For (I = 0; I <n; I ++)
{
Tmp = sa [rank [I]-1];
For (; r [I + k] = r [tmp + k]; k ++)
;
Height [rank [I] = k;
K? -- K: 0;
}
}
Char s [M];
Int solve (int len, int k)
{
Int I;
Node tmp;
Int top = 0, tail = 0;
Height [len + 1] = 0;
If (k = 1)
{
St = 0, ed = len;
Return 1;
}
For (I = 1; I <= len + 1; I ++)
{
Tmp. h = 0;
Tmp. st = I;
While (top <tail & q [tail-1]. w> = height [I])
{
Tmp. h + = q [tail-1]. h;
Tmp. st = q [tail-1]. st;
Tmp. w = q [tail-1]. w;
If (tmp. h = K-1 & tmp. w> height [I])
{
St = sa [tmp. st];
If (top <tail-1)
Ed = st + max (height [I] + 1, q [tail-2]. w + 1 );
Else
Ed = st + height [I] + 1;
Return 1;
}
Tail --;
}
If (height [I] = 0)
Continue;
Tmp. w = height [I];
Tmp. h + = 1;
Q [tail ++] = tmp;
}
Return 0;
}
Int main ()
{
Int I, k, len;
While (scanf ("% d", & k )! = EOF)
{
Scanf ("% s", s );
Len = strlen (s );
For (I = 0; I <len; I ++)
R [I] = s [I] + 1;
R [len] = 0;
Da (len + 1,200 );
Calh (len );
If (solve (len, k) = 0)
{
Puts ("impossible ");
Continue;
}
For (I = st; I <ed; I ++)
Printf ("% c", s [I]);
Puts ("");
}
}