洛谷P3147 [USACO16OPEN]262144

2021年09月15日 阅读数：3

P3147 [USACO16OPEN]262144

题目描述

Bessie likes downloading games to play on her cell phone, even though she doesfind the small touch screen rather cumbersome to use with her large hooves.ide

She is particularly intrigued by the current game she is playing.The game starts with a sequence of NN positive integers (2 \leq N\leq 262,1442N262,144), each in the range 1 \ldots 40140. In one move, Bessiecan take two adjacent numbers with equal values and replace them asingle number of value one greater (e.g., she might replace twoadjacent 7s with an 8). The goal is to maximize the value of thelargest number present in the sequence at the end of the game. Pleasehelp Bessie score as highly as possible!this

Bessie喜欢在手机上下游戏玩（……），然而她蹄子太大，很难在小小的手机屏幕上面操做。spa

输入输出格式

The first line of input contains NN, and the next NN lines give the sequenceinput

of NN numbers at the start of the game.it

Please output the largest integer Bessie can generate.io

输入输出样例

4
1
1
1
2

3

说明

In this example shown here, Bessie first merges the second and third 1s toclass

obtain the sequence 1 2 2, and then she merges the 2s into a 3. Note that it isim

not optimal to join the first two 1s.next

/*
一种 作法相似倍增 动归 递推
f[i][j] 表示 从第i--f[i][j]-1 位这几位的和是 j
而后就能够递推 或者说动归了，由于要相同j是由先后两个相同 j-1合并而来的，
f[i][j]=f[f[i][j-1]][j-1] 合并固然是从小到大合并的
由于先把小的合并了，大的合并机会才会更多嘛，另外一个理由就是根据公式，
在算 j 时 须要事先算出 i-1 才行
*/
#include <cstdio>
#define max(a,b) a>b?a:b
int x,n,ans=0;
int f[60][270000];
int main(){
scanf("%d",&n);
for (int i=1;i<=n;++i)
scanf("%d",&x),f[x][i]=i+1;
for (int i=2;i<=58;++i)
for (int j=1;j<=n;++j){
if (!f[i][j]) f[i][j]=f[i-1][f[i-1][j]];
if (f[i][j]) ans=i;
}
printf("%d\n",ans);
}