#贪心+等差/比数列<序列><嚎叫响彻在贪婪的厂房>

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<map>
 4 #define MAXN 100010
 5 #define ll long long
 6 #define maxn(a,b) (a)>(b)?(a):(b)
 7 #define minn(a,b) (a)<(b)?(a):(b)
 8 using namespace std;
 9 inline ll read(){
10     ll s=0;
11     char ch=getchar();
12     while(ch<‘0‘||ch>‘9‘)ch=getchar();
13     while(ch>=‘0‘&&ch<=‘9‘)s=s*10+ch-‘0‘,ch=getchar();
14     return s;
15 }
16 map<ll,bool>h;
17 ll dp[MAXN],g[MAXN];
18 ll rn[MAXN],n;
19 ll ans=1;
20 inline ll gcd(ll a,ll b){
21     return b?gcd(b,a%b):a;
22 }
23 inline ll get_g(ll a,ll b){
24     if(a==b)return false;
25     if(!a||!b)return false;
26     ll lin;
27     if(a<b)swap(a,b);
28     lin=a/b;
29     if(lin*b!=a)return false;
30     return lin;
31 }
32 bool pd(ll a,ll b){
33     if(a==b)return true;
34     if(!get_g(a,b))return false;
35     ll tmp;
36     while(1){
37         if(a<b)swap(a,b);
38         if(b==1)return true;
39         if(gcd(a,b)!=b){
40             return false;
41         }
42         a/=b;
43     }
44     return true;
45 }
46 int main(){
47 //    freopen("da.in","r",stdin);
48     n=read();
49     for(int i=1;i<=n;++i)rn[i]=read();
50     ans=1;
51     ll llen=1;
52     bool cong=0;
53     for(int i=1;i<=n;++i){
54         if(!cong){
55             llen=1;
56             cong=1;
57             continue;
58         }
59         if(rn[i]!=rn[i-1]){
60             ans=maxn(ans,llen);
61             cong=0;
62             llen=0;
63             --i;
64             continue;
65         }
66         ++llen;
67         ans=maxn(llen,ans);
68     }
69     ans=maxn(ans,llen);
70     for(int i=1;i<n;++i){
71         h.clear();
72         int r=i+1;
73         g[r]=get_g(rn[r-1],rn[r]);
74         if(!g[r]){
75             continue;
76         }
77         dp[r]=2;
78         h[rn[r]]=true;
79         h[rn[r-1]]=true;
80         ans=maxn(ans,dp[r]);
81         for(r=i+2;r<=n;++r){
82             g[r]=get_g(rn[r-1],rn[r]);
83             if(h[rn[r]])break;
84             if(!g[r]){
85                 break;
86             }
87             if(!pd(g[r],g[r-1])){
88                 break;
89             }
90             h[rn[r]]=true;
91             g[r]=maxn(g[r],g[r-1]);
92             dp[r]=dp[r-1]+1;
93             ans=maxn(dp[r],ans);
94         }
95     }
96     printf("%lld\n",ans);
97 }

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<map>
 4 #include<cmath>
 5 #include<iostream>
 6 #define reg register
 7 #define INF 0x7fffffff
 8 #define int long long
 9 using namespace std;
10 inline int minn(int a,int b){return a<b?a:b; }
11 inline int maxn(int a,int b){return a>b?a:b; }
12 inline int read(){
13     int s=0,w=0;char ch=getchar();
14     while(ch<‘0‘||ch>‘9‘)w|=(ch==‘-‘),ch=getchar();
15     while(ch>=‘0‘&&ch<=‘9‘)s=s*10+ch-‘0‘,ch=getchar();
16     return w?-s:s;
17 }
18 #define kd (read())
19 const int MAXN=100010;
20 int N;
21 int rn[MAXN];
22 int gcd(int a,int b){return b?gcd(b,a%b):a; }
23 int ans=0;
24 map<int ,bool >mp;
25 signed main(){
26     //freopen("da.in","r",stdin);
27     N=kd;
28     for(reg int i=1;i<=N;++i)rn[i]=kd;
29     bool xj=1;
30     int gx=0;
31     int pos=1;
32     while(pos<=N){
33         if(xj){
34             mp.clear();
35             ++ans;
36             if(pos==N)break;
37             if(rn[pos]!=rn[pos+1]&&abs(rn[pos]-rn[pos+1])!=1){
38                 gx=maxn(rn[pos+1]-rn[pos],rn[pos]-rn[pos+1]);
39                 mp[rn[pos]]=1;
40                 mp[rn[pos+1]]=1;
41                 xj=0;pos=pos+2;
42             }
43             else xj=1,++pos;
44         }
45         else{
46             int tt=gcd(maxn(rn[pos]-rn[pos-1],rn[pos-1]-rn[pos]),gx);
47             if(tt>1&&!mp[rn[pos]])gx=tt,mp[rn[pos]]=1,++pos;
48             else xj=1;
49         }
50     }
51     printf("%lld\n",ans);
52 }

先丢两个代码，两道题还是很像的。序列里的一个点肯定是尽量和其他的合并，如果考虑从前往后扫，那么这个点能和前面的合并（条件满足情况下）一定不吃亏，所以硬扫O（N）。

不过，细节还是挺多的，比如数列中一定不能有相同的数。还有<嚎叫响彻在贪婪的厂房>中对公差大于1的要求

while(pos<=N){
		if(xj){
			mp.clear();
			++ans;
			if(pos==N)break;
			if(rn[pos]!=rn[pos+1]&&/*/*/abs(rn[pos]-rn[pos+1])!=1/*/*/){//判公差不为1
				gx=maxn(rn[pos+1]-rn[pos],rn[pos]-rn[pos+1]);
				mp[rn[pos]]=1;
				mp[rn[pos+1]]=1;
				xj=0;pos=pos+2;
			}
			else xj=1,++pos;
		}
		else{
			int tt=gcd(maxn(rn[pos]-rn[pos-1],rn[pos-1]-rn[pos]),gx);
			if(tt>1&&!mp[rn[pos]])gx=tt,mp[rn[pos]]=1,++pos;
			else xj=1;
		}
	}

 不够细心鸭！

原文地址：https://www.cnblogs.com/2018hzoicyf/p/11376163.html