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POJ 3273 Monthly Expense 二分答案

2024-07-04 00:11:29   225人阅读

Monthly Expense
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 13281 Accepted: 5362

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 ≤ 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.

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

7 5
100
400
300
100
500
101
400

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.

题解

又是一个二分题。题目的意思是说，在接下来的n天里，Farmer John每天需要花money[i]钱，把这些天分成k份（每份都是连续的天），要求每份的和尽量少，输出这个最小的和。

依旧是二分答案二分答案。。。但是特别奇怪，如果用一个res维护当前可行值就会WA，而且R也得用r = mid 而不是r = mid - 1。。。

这个二分真是难

代码示例

/****
	*@author    Shen
	*@title     poj 3273
	*/

#include <cstdio>
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;

int n, k;
int r, v[100005];
int maxa = 0, mina = 0;

bool test(int x)
{
    int sum = 0, cnt = 1;
    for (int i = 0; i < n; i++)
    {
        if (sum + v[i] <= x)
            sum += v[i];
        else
            cnt++, sum = v[i];
    }
    //printf("\t%s with x = %d, result is that sum = %d.\n", __func__, x, sum);
    return cnt <= k;
}

int Bsearch(int l, int r)
{
    while (l < r)
    {
        int mid = (r + l) / 2;
        //printf("l = %d, r = %d, mid = %d.\n", l, r, mid);
        if (test(mid))
            r = mid;
        else
            l = mid + 1;
    }
    return l;
}

void solve()
{
    maxa = mina = 0;
    for (int i = 0; i < n; i++)
    {
        scanf("%d", &v[i]);
        maxa += v[i];
        mina = max(mina, v[i]);
    }
    int ans = Bsearch(mina, maxa);
    printf("%d\n", ans);
}

int main()
{
    while (~scanf("%d%d", &n, &k))
        solve();
    return 0;
}


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