标签：
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.
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).
Print a single number — the maximum length of an almost constant range of the given sequence.
5
1 2 3 3 2
4
11
5 4 5 5 6 7 8 8 8 7 6
5
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].
我有点惊讶。。B题就考dp了，虽然是简单的dp
题意是说给一个序列，保证相邻两项差值不超过1，求一个最长子串长度，要求满足子串中最大值减最小值小于2
意思就是串中只能有相邻的两个数字咯
令f[i][1]表示以第i个数开头，只包含a[i]和a[i]+1两种数字的最长子串
令f[i][2]表示以第i个数开头，只包含a[i]和a[i]-1两种数字的最长子串
然后
a[i]==a[i+1] 则 f[i][1]=f[i+1][1] f[i][2]=f[i+1][2]
a[i]==a[i+1]+1 则 f[i][1]=1 f[i][2]=f[i+1][1]+1
a[i]==a[i+1]-1 则 f[i][2]=1 f[i][1]=f[i+1][2]+1
1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 #include<cstdlib> 5 #include<algorithm> 6 #include<cmath> 7 #include<queue> 8 #include<deque> 9 #include<set> 10 #include<map> 11 #include<ctime> 12 #define LL long long 13 #define inf 0x7ffffff 14 #define pa pair<int,int> 15 #define pi 3.1415926535897932384626433832795028841971 16 using namespace std; 17 inline LL read() 18 { 19 LL x=0,f=1;char ch=getchar(); 20 while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} 21 while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} 22 return x*f; 23 } 24 inline void write(LL a) 25 { 26 if (a<0){printf("-");a=-a;} 27 if (a>=10)write(a/10); 28 putchar(a%10+‘0‘); 29 } 30 inline void writeln(LL a){write(a);printf("\n");} 31 int n,ans; 32 int a[100010]; 33 int s1[100010]; 34 int s2[100010]; 35 int main() 36 { 37 n=read(); 38 for (int i=1;i<=n;i++)a[i]=read(); 39 s1[n]=s2[n]=ans=1; 40 for (int i=n-1;i>=1;i--) 41 { 42 if (a[i]==a[i+1])s1[i]=s1[i+1]+1,s2[i]=s2[i+1]+1; 43 if (a[i]>a[i+1])s1[i]=1,s2[i]=s1[i+1]+1; 44 if (a[i]<a[i+1])s2[i]=1,s1[i]=s2[i+1]+1; 45 ans=max(ans,max(s1[i],s2[i])); 46 } 47 printf("%d\n",ans); 48 }
cf602B Approximating a Constant Range
标签：
原文地址：http://www.cnblogs.com/zhber/p/5745935.html
