poj 3273（原来以为是dp呢，居然是二分）

最小化月度支出预算

最新推荐文章于 2020-05-28 14:18:17 发布

原创最新推荐文章于 2020-05-28 14:18:17 发布 · 564 阅读

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Farmer John 使用会计技巧管理农场财务，通过合理安排连续天数的支出，以确定最经济的月度预算上限。

Monthly Expense

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14719		Accepted: 5895

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

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

代码很清楚，相信很容易就懂了。

AC代码：

#include<iostream>
using namespace std;
int a[100010];
int main(){
    int n,m;
    int low,high;
    while(cin>>n>>m){
        low=0; high=0;
        for(int i=0;i<n;i++){
            cin>>a[i];
            high+=a[i];
            if(a[i]>low)
                low=a[i];
        }
        while(low<high){
            int mid=(low+high)/2;
            int s,t;
            s=t=0;
            for(int i=0;i<n;i++){
                if(s+a[i]<=mid){
                    s+=a[i];
                }
                else{
                    s=a[i];
                    t++;
                }
            }
            if(t<=m)
                high=mid-1;
            else
                low=mid+1;
        }
        cout<<low<<endl;
    }
    return 0;
}