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…

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…

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…

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…

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 题意:把n个数分成m份,使每份的和尽量小,输出最大的那一个的和. 思路:二分枚举最大的和,时间复杂度为O(nlog(sum-max)); 一道很好的题. #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> using namespace s…

题意:给出n天的花费,需要将这n天的花费分成m组,使得每份的和尽量小,求出这个最小的和 看题目看了好久不懂题意,最后还是看了题解 二分答案,上界为这n天花费的总和,下界为这n天里面花费最多的那一天 如果mid>=m,说明mid偏小,l=mid+1, 如果mid<m,说明mid偏大,r=mid, #include<iostream> #include<cstdio> #include<cstring> #include <cmath> #inclu…

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 ≤…

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…

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(…