对于第一问二分然后贪心判断即可

对于第二问，设f[i][j]为已经到j为止砍了i段，转移的话从$$ f[i][j]=\sigema f[k][j-1] (s[j]-s[k-1]<=ans)

这里用权和嘴和优化成nm的即可

#include<iostream>

#include<cstdio>

using namespace std;

const int N=50005,mod=10007;

int n,m,a[N],f[N],la[N];

long long sm[N],s[N];

int read()

{

	int r=0,f=1;

	char p=getchar();

	while(p>'9'||p<'0')

	{

		if(p=='-')

			f=-1;

		p=getchar();

	}

	while(p>='0'&&p<='9')

	{

		r=r*10+p-48;

		p=getchar();

	}

	return r*f;

}

bool ok(int w)

{

	int sum=0,s=1;

	for(int i=1;i<=n;i++)

	{

		if(sum+a[i]>w)

			sum=a[i],s++;

		else

			sum+=a[i];

	}

	return s<=m;

}

int main()

{

	n=read(),m=read()+1;

	int l=0,r=0,ans,ans2=0;

	for(int i=1;i<=n;i++)

		a[i]=read(),l=max(l,a[i]),r+=a[i],sm[i]=sm[i-1]+a[i];

	while(l<=r)

	{

		int mid=(l+r)>>1;

		if(ok(mid))

			r=mid-1,ans=mid;

		else

			l=mid+1;

	}

	int las=0;

	for(int i=1;i<=n;i++)

	{

		while(sm[i]-sm[las]>ans)

			las++;

		la[i]=las;

	}

	for(int i=0;i<=n;i++)

		s[i]=1;

	for(int i=1;i<=m;i++)

	{

		for(int j=1;j<=n;j++)

			f[j]=(s[j-1]-s[la[j]-1])%mod;

		s[0]=0;

		for(int j=1;j<=n;j++)

			s[j]=s[j-1]+f[j];

		ans2=(ans2+f[n])%mod;

	}

	printf("%d %d\n",ans,ans2);

	return 0;

}