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…
忘开long long了居然没WA 二分答案,枚举判断看最后需要的月份数是否小于等于要求的即可 #include<iostream> #include<cstdio> using namespace std; const int N=100005; int n,m,a[N],l,r,ans; int read() { int r=0,f=1; char p=getchar(); while(p>'9'||p<'0') { if(p=='-') f=-1; p=getch…
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: 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…
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…
二分答案就可以了.... ----------------------------------------------------------------------- #include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<cctype> using namespace std; typedef long long ll; …
一道水题WA了这么多次真是.... 统考终于完 ( 挂 ) 了...可以好好写题了... 先floyd跑出各个点的最短路 , 然后二分答案 m , 再建图. 每个 farm 拆成一个 cow 点和一个 shelter 点, 然后对于每个 farm x : S -> cow( x ) = cow( x ) 数量 , shelter( x ) -> T = shelter( x ) 容量 ; 对于每个dist( u , v ) <= m 的 cow( u ) -> shelter( v…
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…
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…
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…
题目链接:http://poj.org/problem?id=3273 题目大意:给出一个有n个数据的数组,将其分为连续的m份,找到一种分法,是的m份中最大一份总和最小 解题思路: 直接在答案的区间内二分查找,找到符合条件的答案. #include <cstdio> int main() { int n, m; while (scanf("%d %d", &n, &m) != EOF) { int i, j; ]; , right = ; ; i <…
