G - Monthly Expense解题报告（黄杰）

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转载最新推荐文章于 2026-02-27 00:26:56 发布 3.7k 阅读 AI 写同款· GEO 优化 ›

Monthly Expense （最大值最小化＋二分法） Farmer John is an astounding accounting wizard and has realized he might run out of money to run the farm. He has already calculated and recorded the exact amount of money (1 ≤ moneyi ≤ 10,000) that 阅读详情

G - Monthly Expense

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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Description

Farmer John is an astounding accounting wizard and has realized he might run out of money to run the farm. He has already calculated and recorded the exact amount of money (1 ≤ money_i ≤ 10,000) that he will need to spend each day over the next N (1 ≤ N ≤ 100,000) days.

FJ wants to create a budget for a sequential set of exactly M (1 ≤ M ≤ N) fiscal periods called "fajomonths". Each of these fajomonths contains a set of 1 or more consecutive days. Every day is contained in exactly one fajomonth.

FJ's goal is to arrange the fajomonths so as to minimize the expenses of the fajomonth with the highest spending and thus determine his monthly spending limit.

Input

Line 1: Two space-separated integers: N and M
Lines 2.. N+1: Line i+1 contains the number of dollars Farmer John spends on the ith day

Output

Line 1: The smallest possible monthly limit Farmer John can afford to live with.

Sample Input

Sample Output

Hint

If Farmer John schedules the months so that the first two days are a month, the third and fourth are a month, and the last three are their own months, he spends at most $500 in any month. Any other method of scheduling gives a larger minimum monthly limit.

题目大意：FJ有n天，每天的金钱花费为moneyi （每天不一定相同），给定一个m，代表FJ想要创建的周期个数，每个周期包括一天或者连续几天，要求我们计算出最小的周期金钱，这个数大于或等于任何一个周期金钱的和。
题目链接：http://poj.org/problem?id=3273
方法：二分法
思路：二分地量就不说了，都懂得，关键在于判定，可以通过你模拟的值（即二分时的量）来进行周期划分，看满足你模拟的值最少周期数是多少，如果大于
m,显然是不满足的，如果小于，那么划分m组是绰绰有余的，等于就不用说了。。重要的是注意确定一个分段点后，起点在哪儿，是最后一个满足的数？还是接下来这个不满足的数？而且要注意比较一天的情况，因为一天也可能是一个周期。代码有详细的标注。
算法实现：

#include<stdio.h>
#define MAX 100003
int main()
{
	int N,M;
	int low,high,mid;
	int i,res,mon,sum,days,a[MAX];//res记录合适的周期金钱，mon记录分了多少个周期，sum记录连续几天的金钱和，days记录连续几天是可行的，mon记录间断点。
	while(scanf("%d%d",&N,&M)!=EOF)
	{
		high=0;
		for(i=0;i<N;i++)
		{
			scanf("%d",&a[i]);
			high+=a[i];
		}
		low=0;
		while(high-low>=0)
		{
			sum=0;
			mon=0;
			days=0;
			mid=(high+low)/2;
			for(i=0;i<N;i++)
			{
				if(a[i]+sum>mid)
				{
					sum=0;
					if(days>=1)//days是用来分别几天和0天的情况，即是否存在至少一天满足要求。
					{
						i--;//当出现间断点时，即此时间断点在i和i-1之间，a[i]是不符合要求的，所以下一轮要作为起点，所以i--.
						days=0;//且days要清0.
						mon++;//间断点加一。
					}
					else
					{
						mon=-1;//如果只一天得金钱都不符合，则赋值-1，并跳出。
						break;
					}
				}
				else//如果仍满足，sum则累加，连续的天数加一。
				{
					sum+=a[i];
					days++;
				}
				if(mon==-1)//如果有一天的金钱比mid大，则退出，不需在循环。
					break;
			}
			if(mon==-1)
			{
				low=mid+1;
				continue;
			}
			if(mon+1>M)//因为mon是间断点，所以是将整个天数分为mon+1块。
				low=mid+1;
			else
			{
				high=mid-1;
				res=mid;//符合则记录。
			}
		}
		printf("%d\n",res);
	}
	return 0;
}