Monthly Expense
Time Limit: 2000MS |
|
Memory Limit: 65536K |
Total Submissions: 16076 |
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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