poj3273--Monthly Expense(二分)

Monthly Expense

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

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Appoint description:   System Crawler  (2015-01-23)

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

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.

 

题目大意：给出n天的花费，分成m组，要求分的组中的最大值最小。

最大值最小的问题，可以采用二分来解决，二分其中一个值，算另一个结果，找到上届或下届。

在这个题中 二分组的最大花费，计算是不是可以分为m组，找出最小的花费。

 

 

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;
#define LL __int64
LL c[110000] ;
LL n , m ;
int solve(LL k)
{
    int num = 0 , i , sum = 0 ;
    for(i = 0 ; i < n ; i++)
    {
        if( c[i] > k ) return 0 ;
        if( sum+c[i] > k )
        {
            num++ ;
            sum = c[i] ;
        }
        else
            sum += c[i] ;
    }
    if( sum <= k ) num++ ;
    if( num <= m )
        return 1 ;
    return 0;
}
int main()
{
    LL i , j , low , mid , high , last ;
    while( scanf("%I64d %I64d", &n, &m) != EOF )
    {
        low = high = 0 ;
        for(i = 0 ; i < n ; i++)
        {
            scanf("%I64d", &c[i]) ;
            low = max(low,c[i]) ;
            high += c[i] ;
        }
        while( low <= high )
        {
            mid = ( low + high ) / 2 ;
            if( solve(mid) )
            {
                last = mid ;
                high = mid -1 ;
            }
            else
                low = mid + 1 ;
        }
        printf("%I64d\n", last) ;
    }
    return 0;
}