#1,#2,#3,#5WA
查看原帖
#1,#2,#3,#5WA
1071907
lzhgrq楼主2024/8/29 09:26

用二分答案写的,个人认为没问题QWQ

#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<queue>
#include<vector>
#include<stack>
#include<ctime>
#include<cstdlib>
//#include<cmath>
using namespace std;
const int constant=1e5+5;
int n,m,a[constant],mx=-0x7fffffff;
bool _(int mid){
	int cnt=1,num=0;
	for(int i=1;i<=n;i++){
		if(a[i]+num>mid){
			cnt++;
			num=a[i];
		}
		else{
			num+=a[i];
		}
		if(cnt>m||a[i]>mid) return 0;
		
	}
	return 1;
}
int main(){
//	freopen("GRQ.in","r",stdin);
//	freoprn("GRQ.out","w",stdout);
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++){scanf("%d",&a[i]),mx=max(mx,a[i]);}
	int l=0,r=mx;
	while(l<=r){
		int mid=l+(r-l)/2;
		if(_(mid)){
			r=mid-1;
		} 
		else{
			l=mid+1;
		}
	}
	printf("%d",r+1);
	return 0;
}
'''
2024/8/29 09:26
加载中...