POJ 3273 Monthly Expense

Monthly Expense

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 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 5100400300100500101400

Sample Output

500

题目大意：给出N个数，将这些数分为M组，使得每一组的和都不超过某数，求使得分组条件成立的最小的数。

解题思路：二分搜索，注意分组时的判定，当大于mid时返回假，等于时返回真，小于mid时还要将下一个或几个数与当前数的和再与mid作比较。

代码如下：

#include <cstdio>const int maxn = 1000005;const int INF = 1000000;int M,N;int money[maxn];bool judge(int m){    int temp = 0;    int cnt = 0;    for(int i = 0;i < N;i++){        if(money[i] > m)            return false;        temp += money[i];        if(temp > m){            cnt++;            temp = money[i];        }    }    return cnt < M;}int main(){    int lb,ub;while(scanf("%d %d",&N,&M) != EOF){        for(int i = 0;i < N;i++){            scanf("%d",&money[i]);            if(ub < money[i])            ub = money[i];        }        lb = 0;        while(lb <= ub){            int mid = (lb + ub) / 2;            if(judge(mid))                ub = mid - 1;            else                lb = mid + 1;        }        printf("%d\n",lb);    }    return 0;}