Pjudge #21613. 删数/International Zhautykov Olympiad 2022, Computer Science, Day 1, Problem 1.

题面传送门
受到NOIP2021T3的影响容易想到差分。发现就是将差分后两个一样的东西并起来。
然后相当于是问你一直合并最后能剩下最少几个。考虑设\(f_{i}\)为以\(i\)结尾最后能剩下几个，直接暴力转移是\(O(n^2)\)的。
发现这是一个类似于倍增的结构，处理出\(g_{i,j}\)表示\(i\)向前\(2^j\)倍在哪里，然后就可以\(O(n\log W)\)转移了。
code:

#include<bits/stdc++.h>
#define I inline
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define abs(x) ((x)>0?(x):-(x))
#define ll long long
#define db double
#define lb long db
#define N (300000+5)
#define M (27)
#define K (200000+5)
#define mod 9248440332
#define Mod (mod-1)
#define eps (1e-9)
#define U unsigned int
#define it iterator
#define Gc() getchar() 
#define Me(x,y) memset(x,y,sizeof(x))
#define Mc(x,y) memcpy(x,y,sizeof(x))
#define d(x,y) (n*(x-1)+(y))
#define R(n) (rand()*rand()%(n)+1)
#define Pc(x) putchar(x)
#define LB lower_bound
#define UB upper_bound
#define PB push_back
using namespace std;
int n,m,k,x,y,z,st[N],H,A[N],B[N],dp[N],T,F[N][40];ll Po[60];
I void Solve(){
	int i,j,h;scanf("%d",&n);for(i=1;i<=n;i++) scanf("%d",&A[i]);for(i=1;i<n;i++) B[i]=A[i]-A[i+1];
	for(i=1;i<=n;i++) for(j=0;j<=35;j++) F[i][j]=-1;Me(dp,0x3f);dp[0]=0;for(i=1;i<n;i++){
		if(B[i]==0){dp[i]=dp[i-1]+(i==1||B[i-1]!=0);continue;}F[i][0]=i;
		for(j=0;~F[i][j];j++) {
			dp[i]=min(dp[i],dp[F[i][j]-1]+1);if(!B[F[i][j]-1]) break;if(1ll*B[i]*B[F[i][j]-1]<0||B[i]*Po[j]%B[F[i][j]-1]) break;
			h=log2(B[i]*Po[j]/B[F[i][j]-1]);//cerr<<i<<' '<<j<<' '<<F[i][j]<<' '<<B[F[i][j]-1]<<' '<<h<<' '<<B[i]<<'\n';
			if(Po[h]^(B[i]*Po[j]/B[F[i][j]-1]))break;
			F[i][j+1]=F[F[i][j]-1][h];
		}
	}printf("%d\n",dp[n-1]+1);
}
int main(){
	//freopen("1.in","r",stdin);freopen("1.out","w",stdout);
	Po[0]=1;for(int i=1;i<=50;i++) Po[i]=Po[i-1]*2;scanf("%d",&T);while(T--) Solve();
}

标签：Zhautykov,int,Science,ll,Pjudge,db,long,mod,define
来源： https://www.cnblogs.com/275307894a/p/16096907.html