POJ

Monthly Expense

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

题意:给出n天的花费，分m次给，每次给的钱可是一天的花费或者连续几天的花费。最后使每次给的钱的最大值最小化，也就是每次给的钱尽可能的少，求m次中的最大值。

思路:二分答案，最好情况为m==n，最坏情况m=1。所以二分范围为max(daycost[i])~sum。

#include <iostream>#include <fstream>#include <cstdio>#include <cstring>#include <queue>#include <stack>#include <vector>#include <map>#include <cmath>#include <algorithm>#include <functional>#define inf 0X3f3f3f3fusing namespace std;typedef long long ll;const int MAXN=1e9+10;const int MAX=1e6+10;ll num[MAX],maxx,sum;int n,m;bool juge(ll mid){    ll count=0;    int cnt=1;    for(int i=1;i<=n;i++){        count+=num[i];        if(count>mid){            count=0;            i--;            cnt++;        }    }    if(cnt<=m)        return 1;    else        return 0;}int main(){    #ifdef ONLINE_JUDGE    #else    freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    #endif    while(~scanf("%d%d",&n,&m)){        sum=0;maxx=-1;        for(int i=1;i<=n;i++){            scanf("%lld",&num[i]);            maxx=max(maxx,num[i]);            sum+=num[i];        }        ll l=maxx,r=sum,mid;    //二分答案        while(l<r){            mid=(l+r)/2;            if(juge(mid))                r=mid-1;//mid太大            else                l=mid+1;//mid太小        }        cout<<l<<endl;    }    return 0;}