POJ 3273 Monthly Expense【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

500

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
思路:二分,标志性的字眼“最小中的最大值”, “最大之中的最小值”.
很好的解题报告：http://user.qzone.qq.com/289065406/blog/1301655498
代码如下：

View Code

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int n, m, f[100005];
int judge(int mm)
{
    int i, count=0, change=0;
    for(i=0; i<n; i++)
    {
        change+=f[i];
        if(change>mm)
        {
            count++;
            change=f[i];
        }
    }
    count++;
    if(count>m)
        return 1;
    return 0;
}
int main()
{
    while(scanf("%d%d", &n, &m)!=EOF)
    {
        int i, sum=0, maxmon=0;
        memset(f, 0, sizeof(f));
        for(i=0; i<n; i++)
        {
            scanf("%d", &f[i]);
            sum+=f[i];
            maxmon=maxmon>f[i]?maxmon:f[i];
        }
        int left=maxmon, right=sum;
        while(left<right)
        {
            int mid=left+(right-left)/2;
            if(judge(mid))
                left=mid+1;
            else
                right=mid;
        }
        printf("%d\n", left);
    }
}