POJ：3273-Monthly Expense

Monthly Expense

Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 32067 Accepted: 12081

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.

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解题心得：
1. 题意就是给你一个长度为N的序列，现在要让你把他们切割成M份（所以每一份都是连续的），然后每一份都有一个和sum[i]，其中最大的一个是maxSum = max(sum[i]),问这个最大值最小是多少？
2. 最大值最小的问题，二分啊，每次二分枚举然后检查逐步缩小范围。

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#include <algorithm>
#include <stdio.h>
using namespace std;
typedef long long ll;
const int maxn = 1e5+100;
int n,m,num[maxn];

void init() {
    for(int i=0;i<n;i++)
        scanf("%d",&num[i]);
}

bool checke(ll sum) {
    ll temp = 0,cnt = 1;
    for(int i=0;i<n;i++) {
        if(num[i] > sum)
            return false;
        if(temp + num[i] > sum) {
            temp = num[i];
            cnt++;
        } else
            temp += num[i];
    }
    if(cnt <= m)
        return true;
    return false;
}

ll binary_search() {
    ll l = 0,r = 1e9+100;
    while(r-l > 1) {
        ll mid = (l + r) / 2;
        if(checke(mid))
            r = mid;
        else
            l = mid;
    }
    return r;
}

int main() {
    while(scanf("%d%d",&n,&m) != EOF ){
        init();
        ll ans = binary_search();
        printf("%lld
",ans);
    }
    return 0;
}