标签:break false display turn har img clu define 情况下
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!
标签:break false display turn har img clu define 情况下
原文地址:https://www.cnblogs.com/2018hzoicyf/p/11376163.html