Monthly Expense(二分)

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 11196		Accepted: 4587

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

题意简化后的意思就是给出N个数，将这N个数分成M份，每份必须是连续的一个或几个数，要求分的各份的数之和尽可能小，求出M份和中的最大值；
注意，也许在low == high 之前已分成m组，但可能不是最优的，当while(low < high)结束后得到的low肯定是最优结果；

#include<stdio.h>
#include<string.h>
const int N = 100000;
int n,m;//把n个数分成m组；
int money[N+10];
//计算当前mid值能把n个数分成的组数并与m比较；
bool judge(int mid)
{
    int sum = money[0];
    int group = 1;

    for(int i = 1; i < n; i++)
    {
        if(sum + money[i] <= mid)
        {
            sum += money[i];
        }
        else
        {
            group++;
            sum = money[i];
        }
    }

    if(group > m)
        return false;
    else return true;
}

int main()
{
    int low = 0,high = 0;//low为下界，high为上界；
    int mid;
    scanf("%d %d",&n,&m);
    for(int i = 0; i < n; i++)
    {
        scanf("%d",&money[i]);
        high += money[i];//high初始化为n个数的和，相当于把n个数分成一组；
        if(low < money[i])
            low = money[i];//low初始化为n个数中的最大值，相当于把n个数分成n组；
    }

    mid = (low+high)/2;
    while(low < high)
    {
        if(!judge(mid))//如果mid把n个数分得的组数大于m,说明mid偏小，更新low;
        {
            low = mid+1;
        }
        else high = mid-1;//否则说明mid偏大，更新high;
        mid = (low+high)/2;
    }
    printf("%d\n",mid);
    return 0;
}