思路:找最小值最大首先想到二分，我们发现这个时间t对答案来说满足单调性，所以我们二分答案，check时，我们发现就是判断长度为x的区间最小值是否小于t，dp[i]表示1——i-1位置满足题意且i位置必须写题，那么可以得dp[i]=min(dp[j])+w[i]我们用单调队列维护dp[j]即可
#pragma GCC optimize(2)
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include<iostream>
#include<vector>
#include<queue>
#include<map>
#include<stack>
#include<iomanip>
#include<cstring>
#include<time.h>
using namespace std;
typedef long long ll;
#define SIS std::ios::sync_with_stdio(false)
#define space putchar(' ')
#define enter putchar('\n')
#define lson root<<1
#define rson root<<1|1
typedef pair<int,int> PII;
const int mod=1e9+7;
const int N=2e6+10;
const int M=1e5+10;
const int inf=0x3f3f3f3f;
const int maxx=2e5+7;
const double eps=1e-6;
int gcd(int a,int b)
{
return b==0?a:gcd(b,a%b);
}
ll lcm(ll a,ll b)
{
return a*(b/gcd(a,b));
}
template <class T>
void read(T &x)
{
char c;
bool op = 0;
while(c = getchar(), c < '0' || c > '9')
if(c == '-')
op = 1;
x = c - '0';
while(c = getchar(), c >= '0' && c <= '9')
x = x * 10 + c - '0';
if(op)
x = -x;
}
template <class T>
void write(T x)
{
if(x < 0)
x = -x, putchar('-');
if(x >= 10)
write(x / 10);
putchar('0' + x % 10);
}
ll qsm(int a,int b,int p)
{
ll res=1%p;
while(b)
{
if(b&1)
res=res*a%p;
a=1ll*a*a%p;
b>>=1;
}
return res;
}
int n, m;
int dp[N];
int q[N];
int w[N];
bool check(int x){
int hh=0,tt=0;
for(int i=1;i<=n;i++){
if(q[hh]<i-x-1)hh++;
dp[i]=dp[q[hh]]+w[i];
while(hh<=tt&&dp[q[tt]]>=dp[i])tt--;
q[++tt]=i;
}
for(int i=n-x;i<=n;i++)
if(dp[i]<=m)return true;
return false;
}
int main()
{
cin>>n>>m;
int l=0,r=n;
for(int i=1;i<=n;i++)cin>>w[i];
while(l<=r)
{
int mid=l+r>>1;
if(check(mid)) r=mid-1;
else l=mid+1;
}
cout<<r+1<<endl;
return 0;
}