poj3273Monthly Expense 二分

最新推荐文章于 2023-09-01 22:00:12 发布

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本文介绍了一种通过二分查找来解决预算划分的问题，目的是将连续的支出分配到若干个时间段（称为‘fajomonths’），使得任意一个时间段的最大支出达到最小。文章提供了完整的算法思路及C++实现代码。

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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.

题意

1~n天分别有多少钱分m个月看怎么分能让每月花费最大值最小

题解：

二分钱要从钱的最大值开始因为要照顾一下你的check

代码

#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
const int N  = 1e5+100;
int a[N];
int n,m;
bool check(int k){
    int cnt = 1;/*一开始写的0 一直wa QAQ 下面判断的时候就该知道最后一个月没有加上 */
    int p = 0;
    for (int i = 0; i < n; ++i){
        if (a[i]>k) return false;
        p += a[i];
        if (p>k){
            p = a[i];
            cnt++;
        }
    }
    if (cnt<=m) return true;
    return false;
}
int main(){
    scanf("%d%d",&n,&m);
    int mx = -1;
    for (int i = 0; i < n; ++i){
        scanf("%d",&a[i]);
        mx = max(a[i],mx);
    }
    int ld = mx,rd = 1e8;
    while (ld<=rd){
        int mid = (ld+rd)>>1;
        if (check(mid)) rd = mid-1;
        else ld = mid+1;
    }
    printf("%d\n",rd+1);
    return 0;
}