二分:吃巧克力

题目链接:点击这里

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define int long long
#define ll long long
using namespace std;
int n,d;
int left=0;
int right =0;
int a[500005];
int res[500005];
int knoe;
int mid;
int res_cnt=0;
bool check(int x){
    
    
	int sum=0;
	int ans=0;//a数组下标 
	for(int i=1;i<=d;++i){
    
    
    	sum=sum>>1;
    	while(sum<x&&ans<n){
    
    
    		sum+=a[ans++];
		}
		if(sum<x){
    
    
			return false; 
		}
	}
	return true;
}
signed main()
{
    
    
   	scanf("%lld%lld",&n,&d);
   	for(int i=0;i<n;++i){
    
    
   		scanf("%lld",&a[i]);
   		right+=a[i];
	}
	while (left <= right) {
    
    
		//printf("%d %d\n",left,right);
		mid=(left+right)/2;
		if(!check(mid)){
    
    
			right = mid - 1;
		}
		
		else{
    
    
			knoe=mid;
			left=mid+1;
		}
    }
    printf("%lld\n",knoe);
    int sum=0;
    int ans=0;
    
    for(int i=1;i<=d;++i){
    
    
    	sum=sum>>1;
    	while(sum<knoe&&ans<n){
    
    
    		sum+=a[ans++];
			res[res_cnt++]=i;
		}
	}
	while(ans++<n){
    
    
		res[res_cnt++]=d;
	}//吃不完的最后一天一起吃
    for(int i=0;i<res_cnt;++i){
    
    
    	printf("%lld\n",res[i]);
	}
    return 0;
}

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转载自blog.csdn.net/qq_45695839/article/details/108922393