POJ3273Monthly Expense(二分)

Monthly Expense

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 29158 Accepted: 11077

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 ≤ money_i ≤ 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 5100400300100500101400

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.

Source

USACO 2007 March Silver

题目大意:要求将N个数字分成K组，每组经可能小，求K组中最大的值

解题思路:一般这种最大值最小值最好的办法就是用二分，我二分每组大小限制，然后去check在这个范围下是否能够生成K组

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <cmath>#include <cstdlib>#include <ctime>using namespace std;typedef long long LL;int n,k;int a[100005];bool check(int x){int i,flag;int cnt=0;flag=0;for(i=1;i<=n;i++){if(cnt+a[i]>x){flag++;cnt=a[i];}elsecnt+=a[i];}if(flag+1<=k)return true;elsereturn false;}int main(){int l,r,i,sum,ans,mid;cin>>n>>k;sum=0;for(i=1;i<=n;i++){cin>>a[i];sum+=a[i];}l=0,r=sum;while(l<r-1){mid = (l+r)/2;if(check(mid)){r=mid;}else{l=mid;}}sum=0,ans=0;for(i=1;i<=n;i++){if(ans+a[i]>r){sum=max(sum,ans);ans=a[i];}elseans+=a[i];}cout<<sum<<endl;}