POJ_3273Monthly Expense解题报告。

                                                                                                                          Monthly Expense

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 7912 Accepted: 3250

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 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

Line 1: Two space-separated integers: N andM
Lines 2..N+1: Line i+1 contains the number of dollars Farmer John spends on theith 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.

题目连接：http://poj.org/problem?id=3273

算法类型：二分查找。

解题思路：先找出单日消费的最大值，令它为二分查找的的最低端，算出所有钱数，为二分查找的最大值，再进行二份查找，在统计分的个数是否满足题目要求，不满足的话再继续二分。

算法实现：

#include<stdio.h>#define MAX 100001int a[MAX];int N,M;int search(int mid,int m)   //查找函数；{int sum=1,s=0;int i;for(i=0;i<N;i++){s=s+a[i];if(s>mid){ sum++; s=a[i]; if(sum>M) return 0;}}return 1;}int main(){int high=0,low=0,mid;scanf("%d %d",&N,&M);for(int i=0;i<N;i++)    {scanf("%d",&a[i]);high=high+a[i];  //设置上限和下限；if(a[i]>low)low=a[i];}while(low<=high){mid=(high+low)/2;   //二分查找；if(search(mid,M)==0){low=mid+1;}else{high=mid-1;}}printf("%d",mid);return 0;}