/* 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…
                                                                            POJ-3273                                                                                 Monthly Expense Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 10557  …
    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<=moneyi<=10,000),他想要为他连续的M(1<=M<=N)个被叫做“清算月”的结帐时期做一个预算,每一个“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. FJ的目标是安排这些“…
题目链接:http://poj.org/problem?id=3273   Monthly Expense Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 29231   Accepted: 11104 Description Farmer John is an astounding accounting wizard and has realized he might run out of money to run th…
链接:http://poj.org/problem?id=3273 题意:FJ想把n天分成m组,每组是连续的,同一组的花费加起来算,求所分组情况中最高花费的最低值 思路:二分答案.二分整数范围内的花费,每次去check一下,check的过程贪心处理即可. AC代码: #include<iostream> #include<stack> #include<vector> #include<algorithm> #include<cmath> usi…
传送门 https://www.cnblogs.com/violet-acmer/p/9793209.html 题意: 有 N 天,第 i 天会有 a[ i ] 的花费: 将这 N 天分成 M 份,每份包含 1 天或连续的多天: 每份的花费为包含的天数花费的加和,求最大花费的最小值. 题解: 二分搜索答案. AC代码: #include<iostream> #include<cstdio> using namespace std; ; int N,M; int totMoney;…
POJ 3273 Monthly Expense二分查找(最大值最小化问题) 题目: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 ≤ mon…
描述 http://poj.org/problem?id=3273 共n个月,给出每个月的开销.将n个月划分成m个时间段,求m个时间段中开销最大的时间段的最小开销值. Monthly Expense Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 21446   Accepted: 8429 Description Farmer John is an astounding accounting wizard and has…
[POJ 3273] Monthly Expense (二分) 一个农民有块地 他列了个计划表 每天要花多少钱管理 但他想用m个月来管理 就想把这个计划表切割成m个月来完毕 想知道每一个月最少花费多少 每一个月的花费是这个月的花费加和 必须按计划表的顺序来 全部天中花费中最大花费作为下界 全部花费加和作为上界 二分上下界间的花费可能 找出最少每月花费就可以 代码例如以下: #include <iostream> #include <cstdio> using namespace s…
描述 http://poj.org/problem?id=3104 n件衣服,第i件衣服里面有水a[i],自然风干每分钟干1个水,用吹风机每分钟干k个水,但是同时只能对一件衣服使用吹风机,求干完所有衣服所需时间的最小值. Drying Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 12639   Accepted: 3256 Description It is very hard to wash and especial…
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 ≤…
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: 36628 Accepted: 13620 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 reco…
Monthly Expense Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 25959   Accepted: 10021 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…
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…
Monthly Expense Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 19207   Accepted: 7630 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…
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…
题目:http://poj.org/problem?id=3273 二分枚举,据说是经典题,看了题解才做的,暂时还没有完全理解.. #include <stdio.h> #include <string.h> int n, m; ]; bool judge(int x) { , cnt = ; ; i < n; i++) { if(money + a[i] <= x) money += a[i]; else { money = a[i]; cnt++; } } if(c…
Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 11196   Accepted: 4587 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 exac…
//最大值最小 //天数的a[i]值是固定的 不能改变顺序 # include <algorithm> # include <string.h> # include <stdio.h> using namespace std; int n,m; int a[100010]; int judge(int x) { int ans=1;//分成了几组 int tmp=0; for(int i=0;i<n;i++) { tmp+=a[i]; if(tmp>x) {…
https://vjudge.net/problem/POJ-3273 认真审题,代码仔细!!ans的初值应该是1 #include<iostream> #include<cstdio> #include<queue> #include<cstring> #include<algorithm> #include<cmath> #include<map> #define lson l, m, rt<<1 #def…
查看原题 边界,就是边界和思维,怎么有效的判断中间值是大了还是小了,以及准确的找到边界!一个<写成<=就前功尽弃,还特别难找到错误! #include <cstdio> #include <algorithm> ; int N, M; int A[maxN]; using namespace std; int main(void) { == scanf("%d%d", &N, &M)) { , l = , m, sum, cnt, i…
P2884 [USACO07MAR]每月的费用Monthly Expense 二分经典题 二分每个段的限制花费,顺便统计下最大段 注意可以分空段 #include<iostream> #include<cstring> #include<cstdio> using namespace std; int max(int &a,int &b){return a>b?a:b;} #define N 100001 int a[N],n,m,ans=2e9;…
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…
题目:http://poj.org/problem?id=3273 思路:通过定义一个函数bool can(int mid):=划分后最大段和小于等于mid(即划分后所有段和都小于等于mid) 这样我们转化为求 满足该函数的 最小mid.即最小化最大值,可以通过二分搜索来做,要注意二分的边界.WR了好几次. 代码: #include<iostream> #include<string> #include<cstdlib> #include<cstdio> u…
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…
题目链接:cid=80117#problem/E">click here~~ [题目大意] 农夫JF在n天中每天的花费,要求把这n天分作m组.每组的天数必定是连续的.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 [解题思路]: 经典的最小化最大值问题,要求连续的m个子序列,子序列的和最大值的最小,枚举满足条件的m的最小值即为答案.因此二分查找. 1.能否把序列划分为每一个序列之和不大于mid的m个子序列, 2.通过用当前的mid值能把天数分成几组, 3.比較mid和…
Monthly Expense 直接上中文 Descriptions 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? 输入 多组输入输出每组数据第一行是2个整数N,M(1<=M<=N<=100000),接着是N行,每行一个整数vi,表示这个序列. 输出 每组数据输出一行一个数,为这个最大值最小是多少 输入样例 7 510040030010050…
  Copying Books  Before the invention of book-printing, it was very hard to make a copy of a book. All the contents had to be re-written by hand by so called scribers. The scriber had been given a book and after several months he finished its copy. O…
链接:https://www.nowcoder.com/acm/contest/106/K 来源:牛客网 题目描述 It's universally acknowledged that there're innumerable trees in the campus of HUST. Now you're going to walk through a large forest. There is a path consisting of N stones winding its way to…