Monthly Expense
Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d
& %I64u
Submit Status Practice POJ
3273
Appoint description:
System Crawler (2015-05-28)
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
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.
题意：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。
思路：看到各组最小和最大的，果断上二分。很好的一道二分穷举的题。
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <sstream>
#include <algorithm>
#include <set>
#include <queue>
#include <stack>
#include <map>
using namespace std;
typedef long long LL;
const int inf=0x3f3f3f3f;
const double eps=1e-8;
const double pi= acos(-1.0);
const int maxn=100010;
int n,m;
int a[maxn];
int judge(int value)//判断用当前的value值能把天数n分成几组
{
int i;
int sum=0,cnt=1;//开始时把所有的天数分为一组
for(i=0;i<n;i++){//从第一天开始向下遍历每天的花费
if(sum+a[i]<=value)
sum=sum+a[i];
else{
sum=a[i];
cnt++;
}
}
if(cnt>m)//若利用mid值划分的组数比规定的组数要多，则说明mid值偏小
return 0;
else
return 1; //否则mid值偏大
}
int main()
{
int high,low,mid;
while(~scanf("%d %d",&n,&m)){
low=high=0;
for(int i=0;i<n;i++){
scanf("%d",&a[i]);
low=max(low,a[i]);
high+=a[i];
}
while(low<=high){
mid=(low+high)>>1;
if(judge(mid))
high=mid-1;
else
low=mid+1;
}
printf("%d\n",mid);//二分搜索最后得到的mid值必然是使得分组符合要求的最优值
}
return 0;
}