因为没注意到long long 就 TLE 了... 二分一下答案就Ok了.. ------------------------------------------------------------------------------ #include<cstdio> #include<cstring> #include<algorithm> #include<iostream>   #define rep( i , n ) for( int i = 0…

题目链接 Solution 先直接二分答案,然后贪心判断,一旦少于答案就吃一块. 思路很简单,有一点细节. 一天内可以不吃巧克力. 注意处理最后时没吃完的全部在最后一天吃完. Code #include<bits/stdc++.h> #define ll long long #define N 50008 #define inf 0x3f3f3f3f3f3f3f using namespace std; void in(ll &x) { char ch=getchar();ll f=1…

题目 2016: [Usaco2010]Chocolate Eating Time Limit: 10 Sec  Memory Limit: 162 MB Description 贝西从大牛那里收到了N块巧克力.她不想把它们马上吃完,而是打算制定一个计划, 使得在接下来的D天里,她能够尽量地快乐.贝西的快乐指数可以用一个整数来衡量,一开始的时候是0,当她每天晚上睡觉的时候,快乐指数会减半(奇数时向下取整).贝西把她的巧克力按照收到的时间排序,并坚持按照这个顺序来吃巧克力.当她吃掉第i块巧克力的时…

2016: [Usaco2010]Chocolate Eating Time Limit: 10 Sec  Memory Limit: 162 MBSubmit: 224  Solved: 87[Submit][Status][Discuss] Description 贝西从大牛那里收到了N块巧克力.她不想把它们马上吃完,而是打算制定一个计划, 使得在接下来的D天里,她能够尽量地快乐.贝西的快乐指数可以用一个整数来衡量,一开始的时候是0,当她每天晚上睡觉的时候,快乐指数会减半(奇数时向下取整).…

P2985 [USACO10FEB]吃巧克力Chocolate Eating 题目描述 Bessie has received N (1 <= N <= 50,000) chocolates from the bulls, but doesn't want to eat them too quickly, so she wants to plan out her chocolate eating schedule for the next D (1 <= D <= 50,000)…

bzoj2016[Usaco2010]Chocolate Eating 题意: n块巧克力,每次吃可以增加ai点快乐,每天早晨睡觉起来快乐值会减半,求如何使d天睡觉前的最小快乐值最大.n,d≤50000 题解: 二分快乐值,每天不够就吃.注意如果最后一天有剩余巧克力,必须将其全部吃完. 代码: #include <cstdio> #include <cstring> #include <algorithm> #define ll long long #define in…

NC24724 [USACO 2010 Feb S]Chocolate Eating 题目 题目描述 Bessie has received \(N (1 <= N <= 50,000)\) chocolates from the bulls, but doesn't want to eat them too quickly, so she wants to plan out her chocolate eating schedule for the next \(D (1 <= D &…

http://www.lydsy.com/JudgeOnline/problem.php?id=2016 这些最大最小显然是二分. 但是二分细节挺多的...这里注意二分的区间,可以累计所有的可能,然后这就是二分区间的右界..(我是sb) 然后二分的时候,判断那里一定要仔细啊.. 还有这题要开longlong啊(雾) #include <cstdio> #include <cstring> #include <cmath> #include <string>…

二分答案,贪心判断,洛谷上要开long long #include<iostream> #include<cstdio> using namespace std; const int N=50005; int n,m,a[N],b[N]; int read() { int r=0,f=1; char p=getchar(); while(p>'9'||p<'0') { if(p=='-') f=-1; p=getchar(); } while(p>='0'&…

题目链接:https://ac.nowcoder.com/acm/contest/1577/K 题目大意: 给出n块巧克力,m天吃完.每块巧克力有a[i]快乐值,每天可以选择吃任意块或者不吃巧克力(按照给出的巧克力顺序吃),求如何吃巧克力可以使快乐值的最小值最大. 题解思路: 1.二分最小的开心值.check检查是否有方案可以使得每天的快乐值大于当前的mid,若存在,则答案可以增大,即l = mid + 1.若不存在,则答案减小,即r = mid - 1. 2.对于最后得出来的答案,再进行一次g…