文章目录

题目题目分析代码

题目

[题目链接](https://www.luogu.com.cn/problem/P2884)

题目描述 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.

给出农夫在n天中每天的花费，要求把这n天分作m组，每组的天数必然是连续的，要求分得各组的花费之和应该尽可能地小，最后输出各组花费之和中的最大值

输入格式 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

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

输入输出样例 输入 #1复制 7 5 100 400 300 100 500 101 400 输出 #1复制 500 说明/提示 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.

题目分析

一般最大的最小值（最小的最大值），我们二分答案。 二分的下界为每天开支的最大值，上界为所有天花费之和。 mid = （上界+下界）/ 2; check（mid）如果符合就更新答案，把上界变成mid-1，继续二分 否则下界改为mid+1，继续二分。 检查（check）：检查是否可以则检查是否有大于mid的数及分组是否大于m

代码

#include <iostream> using namespace std; const int maxx = 1e5+1; int a[maxx]; int sum = 0,maxn = -1; int n,m; int check(int mid) { int s=0; int coun = m; for(int i = 1;i <= n;i++) { s+=a[i]; if(a[i]>mid) return 0; if(s== mid) { s=0; coun--; } else if(s < mid) { continue; } else { s=a[i]; coun--; } } coun--; if(coun >= 0) return 1; else return 0; } int main() { std::ios::sync_with_stdio(false); cin >> n >> m; int ans; for(int i = 1;i <= n;i++) { cin >> a[i]; sum+=a[i]; //求和作为二分上界 maxn = max(maxn,a[i]); //求最大值作为二分下界 } int l,r; for(l=maxn,r=sum;l<=r;) { int mid = (l+r)/2; if(check(mid)) { ans = mid; r = mid - 1; } else l = mid + 1; } cout << ans << endl; return 0; }

100分~