POJ3273 Monthly Expense

Problem 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 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

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
把所给的n个数分为m分，使得最大的一份总和最小
#include <stdio.h>int a[100000],N,M,H,L,D,i,s,p,t;int f(){    for (p=1,i=s=0;i<N;i++)        if (s+a[i]>D)        {            s=a[i];            p++;        }        else s+=a[i];    if (p>M) return 1;    else return 0;}int main(){int w;//处理整数问题时用一个变量记录当前最佳值，避免向上或向下取整    scanf("%d%d",&N,&M);    scanf("%d",&a[0]);    H=L=a[0];    for (i=1;i<N;i++)    {        scanf("%d",&a[i]);        H+=a[i];        if (a[i]>L) L=a[i];    }    while(H>=L)//此处要加=，因为相等时的w还未记录{D=(H+L)/2;        if (f()) L=D+1;        else {w=D;H=D-1;}    printf("%d",w);    return 0;}
//G++输出用%f,C++用%lf