poj 3273 Monthly Expense (二分)

Monthly Expense

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 34725		Accepted: 12983

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.

 

题意：这题目描述的也是没谁了，完全看不懂。好吧，就是将给定的n个数分成k个区间段，取每种分段方法的区间和的最大值加入一个数组A，这个数组A的最小值就是答案。

 

就是最大化最小值，二分（不懂点这里）

 

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdio>
using namespace std;
int n, k;
int a[100005];
int go(int m)
{
    int h = 0, sum = 1;
    for (int i = 1; i <= n; i++)
    {
        if (h + a[i] > m)
        {
            sum++;
            h = a[i];
        }
        else
            h += a[i];
    }
    if (sum>k)
        return 1;
    else
        return 0;
}

int main()
{
    int sun = 0;
    int l = 0, r;
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i++)
    {
        scanf("%d", &a[i]);
        sun += a[i];
        if (l<a[i])
            l = a[i];//确定下界（n个数里面的最大值）
    }
    r = sun;//上界是所有数的和
    int mid;
    while (l<r)
    {
        mid = (l + r) >> 1;
        if (go(mid))
            l = mid + 1;
        else
            r = mid;
    }
    printf("%d", r);
    return 0;
}