## Solución al problema número 11512 de UVA - 11512.

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``````#include <bits/stdc++.h>
#define INF 0x3F3F3F3F
using namespace std;

typedef pair<int, int> ii;

#define MAX_N 100010                         // second approach: O(n log n)
char T[MAX_N];                   // the input string, up to 100K characters
int n;                                        // the length of input string
int RA[MAX_N], tempRA[MAX_N];        // rank array and temporary rank array
int SA[MAX_N], tempSA[MAX_N];    // suffix array and temporary suffix array
int c[MAX_N];                                    // for counting/radix sort

char P[MAX_N];                  // the pattern string (for string matching)
int m;                                      // the length of pattern string

int Phi[MAX_N];                      // for computing longest common prefix
int PLCP[MAX_N];
int LCP[MAX_N];  // LCP[i] stores the LCP between previous suffix T+SA[i-1]
// and current suffix T+SA[i]
bool cmp(int a, int b) { return strcmp(T + a, T + b) < 0; }      // compare

void countingSort(int k) {                                          // O(n)
int i, sum, maxi = max(300, n);   // up to 255 ASCII chars or length of n
memset(c, 0, sizeof c);                          // clear frequency table
for (i = 0; i < n; i++)       // count the frequency of each integer rank
c[i + k < n ? RA[i + k] : 0]++;
for (i = sum = 0; i < maxi; i++) {
int t = c[i]; c[i] = sum; sum += t;
}
for (i = 0; i < n; i++)          // shuffle the suffix array if necessary
tempSA[c[SA[i]+k < n ? RA[SA[i]+k] : 0]++] = SA[i];
for (i = 0; i < n; i++)                     // update the suffix array SA
SA[i] = tempSA[i];
}

void constructSA() {         // this version can go up to 100000 characters
int i, k, r;
for (i = 0; i < n; i++) RA[i] = T[i];                 // initial rankings
for (i = 0; i < n; i++) SA[i] = i;     // initial SA: {0, 1, 2, ..., n-1}
for (k = 1; k < n; k <<= 1) {       // repeat sorting process log n times
countingSort(k);  // actually radix sort: sort based on the second item
countingSort(0);          // then (stable) sort based on the first item
tempRA[SA[0]] = r = 0;             // re-ranking; start from rank r = 0
for (i = 1; i < n; i++)                    // compare adjacent suffixes
tempRA[SA[i]] = // if same pair => same rank r; otherwise, increase r
(RA[SA[i]] == RA[SA[i-1]] && RA[SA[i]+k] == RA[SA[i-1]+k]) ? r : ++r;
for (i = 0; i < n; i++)                     // update the rank array RA
RA[i] = tempRA[i];
if (RA[SA[n-1]] == n-1) break;               // nice optimization trick
} }

void computeLCP() {
int i, L;
Phi[SA[0]] = -1;                                         // default value
for (i = 1; i < n; i++)                            // compute Phi in O(n)
Phi[SA[i]] = SA[i-1];    // remember which suffix is behind this suffix
for (i = L = 0; i < n; i++) {             // compute Permuted LCP in O(n)
if (Phi[i] == -1) { PLCP[i] = 0; continue; }            // special case
while (T[i + L] == T[Phi[i] + L]) L++;       // L increased max n times
PLCP[i] = L;
L = max(L-1, 0);                             // L decreased max n times
}
for (i = 0; i < n; i++)                            // compute LCP in O(n)
LCP[i] = PLCP[SA[i]];   // put the permuted LCP to the correct position
}

pair<int,pair<int,int>> LRS() {                 // returns a pair (the LRS length and its index)
int i, idx = 1, maxLCP = -1, j=0,flag=0;
for (i = 1; i < n; i++)                         // O(n), start from i = 1
if (LCP[i] > maxLCP) { maxLCP = LCP[i], idx = 2; j=i; flag=1;}
else if (LCP[i] == maxLCP && flag==1) idx++;
else flag=0;
return make_pair(maxLCP,make_pair(j,idx));
}

int main() {
int cases; scanf("%d",&cases);
while(cases--)
{
memset(T, '\0', sizeof(T));
scanf("%s",T);
n = (int)strlen(T);
T[n++] = '\$';
constructSA();                                              // O(n log n)
computeLCP();                                               // O(n)
pair<int,pair<int,int>> ans = LRS();                 // find the LRS of the first input string
pair<int,int> res = ans.second;
string a =  T + SA[res.first];
a = a.substr(0,ans.first);
if(ans.first ==0) printf("No repetitions found!\n");
else printf("%s %d\n", a.c_str(), res.second);
}
return 0;
}
``````

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