Monthly Expense （二分初级典例）

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 ≤ moneyi ≤ 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.

题意：给你n个数，让分成m组，每组必须是连续的几个数，求所有组的和的最大值的最小值。

思路：二分枚举答案。从low（n天中一天的最大花费，理论的最小值）到high（n天的总花费），寻找一个数，检查是否符合题意（以这个数为最大花费，所得最少组数小于m，且low等于high）。用二分就是加快寻找这个数的速度，例如在100内寻找一个数，用遍历的方法平均复杂度是50，用二分就是log2 100，范围越大，二分的速度优势更明显。

#include<iostream>
#include<cmath>
#include<stdio.h>
using namespace std;int n,m;
int a[100002];
bool find(int top){   //检查解是否在mid和low之间int num=1;int cur=0;for(int i=0;i<n;i++){if(cur+a[i]<=top)cur+=a[i];//归入同意组else{num++;    //新的一组cur=a[i];}}return (num<=m);
}int main()
{while(cin>>n>>m){int sum=0;int maxx=0;for(int i=0;i<n;i++){cin>>a[i];sum+=a[i];maxx=max(maxx,a[i]);}int low=maxx,high=sum,mid;//初始化上下界while(low!=high){mid=(high+low)/2;  //二分if(find(mid))high=mid;elselow=mid;}cout<<low<<endl;}return 0;
}

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