NBUT-2014暑期集训专题练习1 -> 二分法L - L

最新推荐文章于 2019-04-20 09:48:00 发布

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农民约翰面临财务困境，已详细记录未来N天每天所需花费。他计划将这些天划分为M个连续日数的组，以确定每月的支出上限，从而最小化最高月支出。

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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.

二分题

#include<stdio.h>
#include<string.h>

int day[100010];
int n,m;
bool _ok(int mid)
{
  int seg=0;
  int sum=0;
  for(int i=0;i<n;i++)
  {
    if(day[i]>mid)//单个就超过mid，显然不行
      return false;
    sum+=day[i];
    if(sum>mid)
    {
      seg++;
      sum=day[i];//多加一组
    }
  }
  return seg<m;
}

int main()
{
  while(~scanf("%d%d",&n,&m))
  {
    int max=0,min=0x3f3f3f3f;
    for(int i=0;i<n;i++)
    {
      scanf("%d",&day[i]);
      max+=day[i];
      if(min>day[i])
        min=day[i];
    }
    int l=min,r=max,mid;
    while(l<=r)
    {
      mid=(l+r)/2;
      if(_ok(mid))
        r=mid-1;
      else
        l=mid+1;
    }
    printf("%d\n",r+1 );
  }
  return 0;
}