题目大意: 共N ( 1 ≤ N ≤ 100,000 )个 工作日 ,分M ( 1 ≤ M ≤ N ) 个 清算月 一个 清算月 包含一个工作日或更多连续的工作日,每一个工作日都仅被包含在一个 清算月 当中. 按顺序分组,得到一个最大值最小化的月度开支(即 在 所有可能的分组结果的最大值 中得到一个最小的) Input Line 1:   Two space-separated integers: N and M Lines 2..N+1:   Line i+1 contains the num…

1639: [Usaco2007 Mar]Monthly Expense 月度开支 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 529  Solved: 269[Submit][Status] Description Farmer John是一个令人惊讶的会计学天才,他已经明白了他可能会花光他的钱,这些钱本来是要维持农场每个月的正常运转的.他已经计算了他以后N(1<=N<=100,000)个工作日中每一天的花费moneyi(1<=money…

直接二分答案然后判断. ----------------------------------------------------------------------------- #include<cstdio> #include<cstring> #include<algorithm> #include<iostream>   #define rep( i , n ) for( int i = 0 ;  i < n ; ++i ) #define c…

1639: [Usaco2007 Mar]Monthly Expense 月度开支 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 593  Solved: 295[Submit][Status] Description Farmer John是一个令人惊讶的会计学天才,他已经明白了他可能会花光他的钱,这些钱本来是要维持农场每个月的正常运转的.他已经计算了他以后N(1<=N<=100,000)个工作日中每一天的花费moneyi(1<=money…

1639: [Usaco2007 Mar]Monthly Expense 月度开支 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 700  Solved: 347[Submit][Status][Discuss] Description Farmer John是一个令人惊讶的会计学天才,他已经明白了他可能会花光他的钱,这些钱本来是要维持农场每个月的正常运转的.他已经计算了他以后N(1<=N<=100,000)个工作日中每一天的花费moneyi(1&…

/* Monthly Expense Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 10757 Accepted: 4390 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 re…

Monthly Expense 题目大意:不废话,最小化最大值 还是直接套模板,不过这次要注意,是最小化最大值,而不是最大化最小值,判断的时候要注意 联动3258 #include <iostream> #include <functional> #include <algorithm> using namespace std; ]; void Search(const int, const int, const int); bool C(const int, cons…

Monthly Expense Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 17982 Accepted: 7190 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 recor…

POJ 3273 Monthly Expense 此题与POJ3258有点类似,一开始把判断条件写错了,wa了两次,二分查找可以有以下两种: ){ mid=(lb+ub)/; if(C(mid)<=m) ub=mid; ; //此时下限过小 } out(ub);//out(lb) 我一开始是写的下面这种,下面这种要单独判断lb和ub的值,因为用下面这种判断lb,ub都可能成立 ){ mid=(lb+ub)/; if(C(mid)<=m) ub=mid; else lb=mid; } if(C(…

Monthly Expense Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 14158   Accepted: 5697 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 r…