Codeforces Round #FF (Div. 2):C. DZY Loves Sequences
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DZY has a sequence a, consisting of n integers.
We'll call a sequence ai, ai + 1, ..., aj(1 ≤ i ≤ j ≤ n) a subsegment of the sequencea. The value (j - i + 1) denotes the length of the subsegment.
Your task is to find the longest subsegment of a, such that it is possible to change at most one number (change one number to any integer you want) from the subsegment to make the subsegment strictly increasing.
You only need to output the length of the subsegment you find.
The first line contains integer n (1 ≤ n ≤ 105). The next line containsn integers a1, a2, ..., an (1 ≤ ai ≤ 109).
In a single line print the answer to the problem — the maximum length of the required subsegment.
67 2 3 1 5 6
5
You can choose subsegment a2, a3, a4, a5, a6 and change its 3rd element (that is a4) to 4.
题意就是, 你可以改变字符串中的一个字符, 就出其最长的连续字串
如案列, 7 2 3 1 5 6 —————7 2 3 4 5 6
输出即为5.
#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<vector>#include<queue>#include<sstream>#include<cmath>using namespace std;#define f1(i, n) for(int i=0; i<n; i++)#define f2(i, m) for(int i=1; i<=m; i++)#define f3(i, n) for(int i=n; i>=1; i--)#define f4(i, n) for(int i=1; i<=n; i++)#define f5(i, n) for(int i=2; i<=n; i++)#define M 1005const int INF = 0x3f3f3f3f;int n, a[100005], b[100005];int main(){ cin>>n; f4(i, n) cin>>a[i]; b[1]=1; f5(i, n) { b[i]=1; if (a[i]>a[i-1]) b[i]=b[i-1]+1; } int ans=-INF; f3(i, n) { if (b[i]==n) ans=max(b[i], ans); else ans=max(ans, b[i]+1); if (a[i-b[i]+1]-1>a[i-b[i]-1]) ans=max(ans,b[i]+b[i-b[i]]); if (a[i-b[i]+2]-1>a[i-b[i]]) ans=max(ans,b[i]+b[i-b[i]]); } cout<<ans<<endl; return 0;}
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