凡是二分题，难点总是在找check，所以只要能找到check，题就很容易了

#include<iostream>
using namespace std;
#define MAX 1000000007
#define N 100007

int n,m,s;
int a[N];

int check(int x){
	int now=0,cns=1;
	for(int i=1;i<=n;i++){
		if(now+a[i]<=x)now+=a[i];
		else {
			cns++;
			i--;
			if(cns>m)return false;
			now=0;
		}
	}
	return true;
}

int search(int l,int r){
	int mid;
	while(l<=r){
		mid=(l+r+1)>>1;
		if(check(mid))r=mid-1;
		else l=mid+1;
	}
	return r+1;
}

int main(){
	ios::sync_with_stdio(false);
	cin>>n>>m;
	for(int i=1;i<=n;i++){
		cin>>a[i];
		s+=a[i];
	}
	
	int ans=search(0,s);
	cout<<ans;
	return 0;
	
}

打卡

#include <bits/stdc++.h>
using namespace std;
int w[100007];
int main(){
	int n,m,right=0,left=0;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++){
		scanf("%d",&w[i]);
		right+=w[i];
		left=max(left,w[i]);
	}
	int mm=right;
	while(left<=right){
		int mid=(left+right)>>1;
		int ans=1;
		int sum=0;
		for(int i=1;i<=n;i++){
			if(sum+w[i]>mid){
				ans++;
				sum=w[i];	
			}
			else sum+=w[i];
		}
		if(ans<=m)  right=mid-1;
		else left=mid+1;
	}
	cout<<right+1;
	return 0;
}