lg3147

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链接:lg3147

题解

  • dp[i][j] 表示第i个位置合成到j所需的长度
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#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<map>
#include<algorithm>
#define rint register int
#define rep(i,x,y) for(rint i=x;i<=y;++i)
#define repd(i,x,y) for(rint i=x;i>=y;--i)
#define lowbit(x) (x&(-x))
#define ll long long
using namespace std;
const int N=1e6;
int dp[N][70],a[N],n,ans;
template <typename T>inline void read(T &x){
	x=0;char c;int sign=1;
	do{c=getchar();if(c=='-')sign=-1;}while(c<'0'||c>'9');
	do{x=x*10+c-'0'; c=getchar();}while(c>='0'&&c<='9');
	x*=sign;
}
int main(){
	read(n);
	rep(i,1,n)read(a[i]);
	rep(i,1,n){
		dp[i][a[i]]=1;
		int j=i-1;
		while(dp[j][a[i]]){
			j-=dp[j][a[i]];
			a[i]++;dp[i][a[i]]=i-j;	
		}
		ans=max(ans,a[i]);
	}
	cout<<ans<<endl;
	return 0;
}