06.04. Monthly Expense [Binary Search & Parametric Search][백준 6236]

https://www.acmicpc.net/problem/6236

6236번: 용돈 관리

 

Algorithm classification : Binary Search, Parametric Search

 

06.04.1. Problem

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.

 

06.04.2. Input limit

• 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

 

06.04.3. Output

• Line 1: The smallest possible monthly limit Farmer John can afford to live with.

 

06.04.4. Solution

최소금액이 구하는 목표이니 이를 MID 값으로 추적해주면 됩니다. 

	int n, m;
	scanf("%d %d", &n, &m);
	int a[n], i, left = 0;
	
	
	for(i = 0; i < n; i++){
		scanf("%d", &a[i]);
		left = max(left, a[i]);
	}
	
	int right = 1000000000, mid, temp, sum, result;
	while(left <= right){
		mid = (left + right) / 2;
		temp = 1;
		sum = 0;
		for(i = 0; i < n ; i++){
			if(sum + a[i] <= mid){
				sum += a[i];
			}
			else{
				sum = a[i];
				temp++;
			}
		}
		//printf("%d %d %d > %d\n",left, right, mid, temp);
		if(temp <= m){
			result = mid;
			right = mid - 1;
		}
		else{
			left = mid + 1;
		}
	}
	
	printf("%d", result);

 

아래에 몇 개의 test case를 남겨두었습니다.

5 4
100
100
100
100
100
>> 200

5 1
1
3
5
7
9
>> 25

5 5
1
1
1
1
100
>> 100