POJ 3273-Monthly Expense（二分法-最小化最高花费）

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

Output

Sample Input

7 5100400300100500101400
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.
Source
USACO 2007 March Silver
题目意思：
给出John在N个月中每个月的花费，John想将N个月分成M个组。每组的月份是连续的，同一组的花费被相加起来，求所有分组情况中最高花费的最低值。
解题思路：
二分法，所求答案ans在[low,high]范围内二分。low是最高的单月花费；high是各个月所有花费之和，mid是当前的ans。遍历各个月份，将花费总和不超过ans的月份归入一个组，统计在当前ans情况下能分成几个组。如果组数少于M，说明ans偏大，范围变成前二分之一；否则说明ans偏小，范围变成后二分之一。
#include<iostream>#include<cstdio>#include<iomanip>#include<cmath>#include<cstdlib>#include<cstring>#include<map>#include<algorithm>using namespace std;#define MAXN 100000int a[MAXN];int n,m,high=0,low=-1,mid;bool solve(){    int res=1,sum=0;    for(int i=0; i<n; ++i)    {        sum+=a[i];//若干月花费总和        if(sum>=mid)//超过限额        {            ++res;//分组数目            if(sum==mid) sum=0;//当前月花费计入当前组            else sum=a[i];//当前月花费计入下一组            if(i==n-1) --res;//最后一个月刚好被计入最后一组        }    }    if(res>m) return false;    else return true;}int main(){#ifdef ONLINE_JUDGE#else    freopen("F:/cb/read.txt","r",stdin);    //freopen("F:/cb/out.txt","w",stdout);#endif    ios::sync_with_stdio(false);    cin.tie(0);    cin>>n>>m;    for(int i=0; i<n; ++i)    {        cin>>a[i];        if(a[i]>low) low=a[i];//最高的单月花费        high+=a[i];//所有花费之和    }    while(low<high)    {        mid=0.5*(low+high);        if(solve()) high=mid-1;        else low=mid+1;    }    cout<<low<<endl;    return 0;}/*7 5100400300100500101400*/