POJ-3273-Monthly Expense

Monthly Expense

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 22175 Accepted: 8681

Description

Farmer John is an astounding(令人震惊的) accounting wizard(男巫) and has realized he might run out of money to run the farm. He has already calculated(计算) and recorded the exact amount(数量) 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(预算) for a sequential(连续的) set of exactly M (1 ≤ M ≤ N) fiscal(会计的) periods called "fajomonths". Each of these fajomonths contains a set of 1 or more consecutive(连贯的)days. Every day is contained in exactly one fajomonth.

FJ's goal is to arrange the fajomonths so as to minimize(使减到最少) the expenses of the fajomonth with the highest spending and thus determine his monthly spending limit.

Input

Line 1: Two space-separated integers(整数): 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

Hint

If Farmer John schedules(时间表) 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(最小的) monthly limit.

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

二分

#include<iostream>#include<string.h>#include<stdio.h>#include<math.h>#include<algorithm>#include<queue>#include<vector>#include<map>using namespace std;int main(){    int n, m, low, high, mid, MAX = 0, a[100010], sum = 0, ans;    scanf("%d%d", &n, &m);    for(int i = 0;i<n;++i)    {        scanf("%d", &a[i]);        sum+=a[i];        if(MAX<a[i])MAX= a[i];    }    high = sum;    low = MAX;    while(low<=high)    {        mid = (low+high)/2;        ans = 1;        sum = 0;        for(int i = 0; i<n;++i)        {            sum+=a[i];            if(sum>mid)            {                sum = a[i];                ans++;            }        }        if(ans<=m)        {            high = mid-1;            MAX = mid;        }        else            low = mid+1;    }    printf("%d\n", MAX);    return 0;}