poj3273 最大值最小化

 

如题：http://poj.org/problem?id=3273

Monthly Expense

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 17781 Accepted: 7115

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 nextN (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 andM
Lines 2..N+1: Line i+1 contains the number of dollars Farmer John spends on theith day

Output

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

Sample Input

7 5100400300100500101400

Sample Output

500

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.

Source

USACO 2007 March Silver

 

 

 

思路：二分钱数，使用函数C判断n天划分成每一份<=x,最少划分的份数是否>m，如果返回1，则在右边搜索，否则左边搜索。

 

 

 

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define MAXN 100005
#define max(a,b)(a>b?a:b)

int a[MAXN];

bool C(int x,int n,int m) //n天划分成每一份<=x,最少划分的份数是否>m
{
 int sum=0;
 int i=0;
 int cnt=0;
 for(i=0;i<n;i++)
 {
  if((sum+a[i])<=x)
   sum+=a[i];
  else if(a[i]>x)
   return 1;
  else
  {
   sum=a[i];
   cnt++;
  }
 }
  cnt++;
 if(cnt>m)
  return 1;
 return 0;
}

int main()
{
 //freopen("C:\\Users\\Administrator\\Desktop\\1\\expense.5.in","r",stdin);
 int N,M;
 int i;
 cin>>N>>M;
 int l=0,r=0;
 for(i=0;i<N;i++)
 {
  cin>>a[i];
  r=max(r,a[i]);
 }
 r=r*M+1;
 while(r-l>1)
 {
  int mid=(l+r)/2;
  if(C(mid,N,M))
   l=mid;
  else
   r=mid;
 }
 cout<<r<<endl;
 return 0;
}