題目分析：

我們用F[I]$F[I]$表示用第IN$IN$包做草堆的時候最底下一層的最短長度，G[I]$G[I]$記錄此時能到達的最高高度，顯然可以得到如下的一個狀態轉移方程：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define DB double
#define SG string
#define LL long long
#define Fp(A,B,C,D) for(A=B;A<=C;A+=D)
#define Fm(A,B,C,D) for(A=B;A>=C;A-=D)
#define Clear(A) memset(A,0,sizeof(A))
 

#define Copy(A,B) memcpy(A,B,sizeof(A))
using namespace std;
const LL Max=1e5+5;
const LL Mod=1e9+7;
const LL Inf=1e18;
LL N,F[Max],G[Max],Q[Max],Sum[Max];
inline LL Read(){
    LL X=0;char CH=getchar();bool F=0;
    while(CH>'9'||CH<'0'){if(CH=='-')F=1;CH=getchar();}
    while(CH>='0'&&CH<='9'){X=(X<<1)+(X<<3)+CH-'0';CH=getchar();}
    return F?-X:X;
}
inline void Write(LL X){
    if(X<0)X=-X,putchar('-');
    if(X>9)Write(X/10);
    putchar(X%10+48);
}
int main(){
    LL I,J,K;
    N=Read();
    Fp(I,1,N,1){
        Sum[I]=Sum[I-1]+Read();
    }LL Head=1,Tail=1;
    Q[1]=N+1;
    Fm(I,N,1,1){
        while(Head<Tail&&Sum[Q[Head+1]-1]-Sum[I-1]>=F[Q[Head+1]]){
            ++Head;
        }F[I]=Sum[Q[Head]-1]-Sum[I-1];G[I]=G[Q[Head]]+1;
        while(Head<=Tail&&Sum[Q[Tail]-1]-F[Q[Tail]]<=Sum[I-1]-F[I]){
            --Tail;
        }Q[++Tail]=I;
    }Write(G[1]);
    return 0;
}