POJ 3273：Monthly Expense 二分好题啊啊啊啊啊啊

Monthly Expense

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 19207		Accepted: 7630

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.

题意是给出一系列数，将这些数按顺序分成M组，求每组和的最小值是多少。

枚举答案，left是一个数一个组，即最大值。right是所有数的和。

每次去和M比较来判断mid是多是少。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

int N,M;
int value[100000];

int solve(int mid)
{
	int i,sum=0,result=1;
	if(mid==501)
		i=1;
	for(i=1;i<=N;i++)
	{
		if(value[i]+sum<=mid)
		{
			sum += value[i];
		}
		else
		{
			sum=value[i];
			result++;
		}
	}
    if(result>M)
	{
		return 1;
	}
	else
		return -1;
}

int main()
{
	int i,sum,max_x;
	cin>>N>>M;

	sum=0;
	max_x=0;
	for(i=1;i<=N;i++)
	{
		cin>>value[i];
		sum += value[i];
		max_x=max(max_x,value[i]);
	}
	
	int mid=(sum+max_x)/2;
	while(max_x<sum)
	{
		int temp=solve(mid);
		if(solve(mid)==-1)
		{
			sum = mid-1;
		}
		else
		{
			max_x = mid+1;
		}
		mid = (max_x+sum)/2;
	}

	cout<<mid<<endl;
	return 0;
}