POJ 3273 二分

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

Sample Output

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：

给出农夫在n天中每天的花费，要求把这n天分作m组，每组的天数必然是连续的，要求分得各组的花费之和应该尽可能地小，最后输出各组花费之和中的最大值。最大化最小值即可，二分搜索。

下见代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn=100005;
int n,m;
int a[maxn];
bool C(int x) //c(x)=能不能把n个数分成最多m个每个组的和不超过x的组。（有点绕，但这是重点，仔细理解）
{
    int sum=0;
    int num=1;
    for(int i=0;i<n;i++)
    {
        sum+=a[i];
        if(sum>x)
        {
            sum=a[i];
            num++;
        }
    }
    return num<=m;
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
         int MAX=0;
         int asum=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            MAX=max(MAX,a[i]);
            asum+=a[i];
        }
        int lb=MAX,ub=asum,ans=0;
        while(lb<=ub)
        {
            int mid=(lb+ub)/2;
            if(C(mid))
            {
                ans=mid;
                ub=mid-1;
            }
            else lb=mid+1;
        }
        printf("%d
",ans);
    }
    return 0;
}