POJ 3273 Monthly Expense

Monthly Expense

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 16076		Accepted: 6408

Description

Farmer John is an astounding Accounting wizard and have 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'll need to spend Each day of 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 the these fajomonths contains a set of 1 or more consecutive days. Every day was 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: Nand 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 5100400300100500101400

Sample Output

500

Hint

If Farmer John Schedules The months so, the first is a month, the third and fourth is a month, and the last Three is their own months, he spends at the most of any month. Any other method of scheduling gives a larger minimum monthly limit.

 #include <iostream> #include <stdio.h> #include <cstring> #include <string> #include <cmath> #include <algorithm> #define N 200009using namespace Std;int n,m;int a[n];int    Fun (int mm) {int num=1;//Note that the total number of days is initially used as a set of int sum=0;        for (int i=1;i<=n;i++) {if (sum+a[i]<=mm) sum+=a[i];            else {sum=a[i];        num++;    }} if (num>m) return 1; else return 0;}        int main () {while (~scanf ("%d%d", &n,&m)) {int mmax=0;        int ss=0;            for (int i=1;i<=n;i++) {scanf ("%d", &a[i]);            Ss+=a[i];        if (Mmax<a[i]) mmax=a[i];        } int le=mmax,ri=ss;        int mid;            while (Le<ri) {mid= (Le+ri)/2;            if (mid) le=mid+1;        else ri=mid-1;    } cout<<le<<endl; } return 0;} 

POJ 3273 Monthly Expense