POJ3273-----Monthly Expense
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题目大意:把n个数分为连续的m分,尽可能是每份最少;
思路:二分最小值;
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 nextN (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
Lines 2..N+1: Line i+1 contains the number of dollars Farmer John spends on theith day
Output
Sample Input
7 5100400300100500101400
Sample Output
500
Hint
Source
第一次自己写的二分过题;激动;
#include <iostream>#include<cstdio>#include<cstring>#include<cmath>#include<queue>#include<algorithm>#define LL long longusing namespace std;const int N= 1e5;LL a[N+10];int m,n;bool solve(LL p){ int i=0,tmp,ans=0; while(i<n) { tmp=a[i]; while(i<n&&tmp<=p) { tmp+=a[++i]; } ans++; if(ans>m) return 0; } return 1;}int main(){ while(~scanf("%d%d",&n,&m)) { for(int i=0;i<n;i++) { scanf("%lld",&a[i]); } LL low=0,high=1e10,mid; LL ans; while(low<=high) { mid=(low+high)/2; if(solve(mid)) { ans=mid; //printf("%d\n",ans); high=mid-1; } else { low=mid+1; } } cout<<ans<<endl; } return 0;}
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