Longest Repeated Sequence

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

描述

You are given a sequence of integers, A = a1, a2, ... an. A consecutive subsequence of A (say ai, ai+1 ... aj) is called a "repeated sequence" if it appears more than once in A (there exists some positive k that ai+k= ai, ai+k+1 = ai+1, ... aj+k = aj) and its appearances are not intersected (i + k > j).

Can you find the longest repeated sequence in A?

输入

Line 1: n (1 <= n <= 300), the length of A.
Line 2: the sequence, a1 a2 ... an (0 <= ai <= 100).

输出

The length of the longest repeated sequence.

样例输入

52 3 2 3 2

样例输出

2

未测试，要注意的是search参数为4个迭代器，而end迭代器要指向范围的下一个元素，故搜索范围为A.begin() + i, A.begin() + i + k+1。

循环条件为i+k<n，条件太弱。

当剩余搜索的序列长度小于要搜索的子序列长度则已不成立，即：

(i+k)-i<=n-(i+k+1)

i+2k+1<=n

#include <iostream>#include <map>#include <vector>#include <string>#include <algorithm>#include <cstring>#include <cstdio>#include <cmath>using namespace std;int main(){int n;cin >> n;vector<int> A(n);for (size_t i = 0; i != n; ++i)cin >> A[i];int len=0, k=0;for (size_t i = 0; i != n; ++i){//while (i + k < n)while (i+2*k+1<=n){//   beg                           end          beg             end!! need + 1if (search(A.begin() + i + k + 1, A.end(), A.begin() + i, A.begin() + i + k+1) != A.end()){len = max(k+1, len);++k;}elsebreak;}k = 0;}cout << len << endl;return 0;}