POJ 3273, Monthly Expense

Time Limit: 2000MS  Memory Limit: 65536K
Total Submissions: 4105  Accepted: 1620

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 ≤ moneyi ≤ 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
7 5
100
400
300
100
500
101
400

 

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

---

//  POJ3273.cpp : Defines the entry point for the console application.
//

#include  < iostream >
#include  < numeric >
#include  < algorithm >
using   namespace  std;

int  main( int  argc,  char *  argv[])
{
     int  N, M;
     int  day[ 100005 ];

    scanf( " %d %d " ,  & N,  & M);
     for ( int  i  =   0 ; i  <  N;  ++ i)
        scanf( " %d " ,  & day[i]);

     int  maxv  =  accumulate( & day[ 0 ], & day[N], 0 );

     int  minv  =   * max_element( & day[ 0 ], & day[N]);

     int  midv;

     while (minv  <  maxv)
    {
        midv  =  (minv  +  maxv) >> 1 ;
         int  k  =   0 ;
         int  sum  =   0 ;
         for  ( int  i  =   0 ; i  <  N;  ++ i)
        {
            sum  +=  day[i];
             if  (sum  >  midv)
            {
                sum  =  day[i];
                 ++ k;
            }
        }

         if  (k  <  M) maxv  =  midv;
         else  minv  =  midv  +   1 ;
    }

    cout  <<  maxv  <<  endl;
     return   0 ;
}

转载于:https://www.cnblogs.com/asuran/archive/2009/10/28/1591017.html