Problem 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

題意：給出n天每天的花費，將n天分成m組，且分組的天數是連續的，要求分完各組的花費和盡量小，輸出各組花費和的最大值。

思路：一開始以為是 dp ，寫了寫沒寫出來突然想到要求各組最小和的最大值應該用二分做，順利AC。。。

Source Program

#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<string> #include<cstdlib> #include<queue> #include<vector> #define INF 0x3f3f3f3f #define PI acos(-1.0) #define N 100001 #define MOD 123 #define E 1e-6 using namespace std; int n,m; int a[N]; bool judge(int value)//判斷當前花費可把n分成幾組 {int sum=0;//花費總和int cnt=1;//初始時只有一組for(int i=0;i<n;i++)//枚舉每天花費{if(sum+a[i]<=value)sum+=a[i];else{sum=a[i];cnt++;}}if(cnt>m)//若分組數比規定數要多，則mid偏小，需調整左值return false;else//若分組數比規定數要少，則mid偏大，需調整右值return true; } int main() {while(scanf("%d%d",&n,&m)!=EOF){int left=0,right=0;for(int i=0;i<n;i++){scanf("%d",&a[i]);left=max(left,a[i]);right+=a[i];}int mid;while(left<=right){mid=(left+right)/2;if(judge(mid))right=mid-1;elseleft=mid+1;}printf("%d\n",mid);}return 0; }

?

總結

以上是生活随笔為你收集整理的Monthly Expense（ POJ-3273 ）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。