Codeforces Round #333 (Div. 2) B. Approximating a Constant Range
2015-11-27 21:31
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B. Approximating a Constant Range
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
When Xellos was doing a practice course in university, he once had to measure the intensity of an effect that slowly approached equilibrium. A good way to determine the equilibrium intensity would be choosing a sufficiently large number of consecutive data
points that seems as constant as possible and taking their average. Of course, with the usual sizes of data, it's nothing challenging — but why not make a similar programming contest problem while we're at it?
You're given a sequence of n data points a1, ..., an.
There aren't any big jumps between consecutive data points — for each 1 ≤ i < n, it's guaranteed that |ai + 1 - ai| ≤ 1.
A range [l, r] of data points is said to be almost constant if
the difference between the largest and the smallest value in that range is at most 1. Formally, let M be
the maximum and m the minimum value of ai for l ≤ i ≤ r;
the range [l, r] is almost constant if M - m ≤ 1.
Find the length of the longest almost constant range.
Input
The first line of the input contains a single integer n (2 ≤ n ≤ 100 000) —
the number of data points.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 100 000).
Output
Print a single number — the maximum length of an almost constant range of the given sequence.
Sample test(s)
input
output
input
output
Note
In the first sample, the longest almost constant range is [2, 5]; its length (the number of data points in it) is 4.
In the second sample, there are three almost constant ranges of length 4: [1, 4], [6, 9] and [7, 10];
the only almost constant range of the maximum length 5 is [6, 10].
题目大意:
给你一串数,让你求最大的区间(最大值-最小值<=1)
解题思路:
我是用集合做的。。。
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
When Xellos was doing a practice course in university, he once had to measure the intensity of an effect that slowly approached equilibrium. A good way to determine the equilibrium intensity would be choosing a sufficiently large number of consecutive data
points that seems as constant as possible and taking their average. Of course, with the usual sizes of data, it's nothing challenging — but why not make a similar programming contest problem while we're at it?
You're given a sequence of n data points a1, ..., an.
There aren't any big jumps between consecutive data points — for each 1 ≤ i < n, it's guaranteed that |ai + 1 - ai| ≤ 1.
A range [l, r] of data points is said to be almost constant if
the difference between the largest and the smallest value in that range is at most 1. Formally, let M be
the maximum and m the minimum value of ai for l ≤ i ≤ r;
the range [l, r] is almost constant if M - m ≤ 1.
Find the length of the longest almost constant range.
Input
The first line of the input contains a single integer n (2 ≤ n ≤ 100 000) —
the number of data points.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 100 000).
Output
Print a single number — the maximum length of an almost constant range of the given sequence.
Sample test(s)
input
5 1 2 3 3 2
output
4
input
11 5 4 5 5 6 7 8 8 8 7 6
output
5
Note
In the first sample, the longest almost constant range is [2, 5]; its length (the number of data points in it) is 4.
In the second sample, there are three almost constant ranges of length 4: [1, 4], [6, 9] and [7, 10];
the only almost constant range of the maximum length 5 is [6, 10].
题目大意:
给你一串数,让你求最大的区间(最大值-最小值<=1)
解题思路:
我是用集合做的。。。
#include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <vector> #include <queue> #include <algorithm> #include <set> using namespace std; #define MM(a) memset(a,0,sizeof(a)) typedef long long LL; typedef unsigned long long ULL; const int maxn = 1e5+5; const int mod = 1e9+7; const double eps = 1e-10; const int INF = 0x3f3f3f3f; LL gcd(LL a, LL b) { if(b == 0) return a; return gcd(b, a%b); } multiset <int> s; int arr[maxn]; int main() { int n; while(cin>>n) { s.clear(); for(int i=0; i<n; i++) scanf("%d",&arr[i]); int i , j, ret = -INF; i = j = 0; while(j < n) { s.insert(arr[j]); if(abs(*s.begin()-*s.rbegin()) > 1) { s.erase(s.find(arr[i])); i++; } ret = max(ret, int(s.size())); j++; } cout<<ret<<endl; } return 0; }
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