POJ3273:Monthly Expense(二分)
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 9302 | Accepted: 3823 |
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 exactlyM (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 5 100 400 300 100 500 101 400
Sample Output
500
Hint
Source
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define MAX 100010
int v[MAX];
int sum;
int n,m;
int most;
bool check(int lim)
{
int i;
int s=0;
int ct=0;
for(i=1;i<=n;i++)
{
if(s+v[i]<=lim)
s+=v[i];
else
{
ct++;
s=v[i];
}
}
ct++;
if(ct<=m)
return true;
return false;
}
int search()
{
int lt=most,rt=sum;
while(lt<rt)
{
int mid=(lt+rt)/2;
if(check(mid))
rt=mid;
else
lt=mid+1;
}
return rt;
}
int main()
{
while(scanf("%d%d",&n,&m)!= EOF)
{
int i;
sum=0;
most=-1;
for(i=1;i<=n;i++)
{
scanf("%d",&v[i]);
sum+=v[i];
if(v[i]>most)
most=v[i];
}
int ans=search();
printf("%d\n",ans);
}
}