目录 题目链接 题解 代码 题目链接 AGC027 A - Candy Distribution Again 题解 贪心即可 代码 #include<cstdio> #include<cstring> #include<algorithm> #define gc getchar() #define pc putchar inline int read() { int x = 0,f = 1; char c = gc; while(c < '0' || c >…

Candy Distribution 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5291 Description WY has n kind of candy, number 1-N, The i-th kind of candy has ai. WY would like to give some of the candy to his teammate Ecry and lasten. To be fair, he hopes that…

Candy Distribution Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 544    Accepted Submission(s): 214   Problem DescriptionWY has n kind of candy, number 1-N, The i-th kind of candy has ai. WY w…

Kids like candies, so much that they start beating each other if the candies are not fairly distributed. So on your next party, you better start thinking before you buy the candies. If there are KK kids, we of course need K⋅XK⋅X candies for a fair di…

可以证明: f(k) = k *(k - 1)/ 2 (1 ≤ k ≤ n)是n的完全剩余系当且仅当n = 2 ^ t. http://poj.org/problem?id=3372…

题目大意:把$x$个糖果分给$n$个人,必须分完,如果第$i$个人拿到$a_i$个糖果,就会开心,输出最多多少人开心 题解:从小到大排序,判断是否可以让他开心,注意最后判断是否要少一个人(没分完) 卡点:无 C++ Code: #include <cstdio> #include <algorithm> #define maxn 111 int n, x, ans; int a[maxn]; int main() { scanf("%d%d", &n,…

Description AGC027A 你有一些糖果,你要把这些糖果一个不剩分给一些熊孩子,但是这帮熊孩子只要特定数目的糖果,否则就会不开心,求最多的开心人数. Solution 如果\(\sum a_i = x\)的话,答案就是\(N\),否则答案一定小于\(N\).对于一个熊孩子们的真子集\(S\),如果\(\sum_{a_i\in S} a_i \le x\),那么一定可以满足这些孩子,然后把剩下的糖果任意分配到剩下的孩子手中即可.这样我们只需要找到最大的\(S\),排序后贪心即可.但是不…

[赛时·8]AGC-027 日常AGC坑……还好能涨Rating +传送门+ ◇ 简单总结 感觉像打多校赛一样,应该多关注一下排名……考试的时候为了避免影响心态,管都没有管排名,就在那里死坑B题.最后还剩十几分钟的时候发现大家都把C题AC了,B题空了一大堆.后来事实也证明了C题比B题简单. ◇ 题目&解析 ◆A题◆ Candy Distribution Again ❄模拟❄ ·[题意] 老师要将x颗糖分给n个孩子(恰好分完).当第i给孩子得到恰好a[i]颗糖时,他会感到快乐.问如何给孩子分糖使得…

Question There are N children standing in a line. Each child is assigned a rating value. You are giving candies to these children subjected to the following requirements: Each child must have at least one candy. Children with a higher rating get more…

A - Candy Distribution Again 大意:有x个糖给n个小朋友,必须分完,小朋友得到糖数为一个确切值的时候小朋友会开心,求最多的开心数 题解 直接排序然后贪心分,如果分到最后一个有剩余那么开心数-1即可 代码 #include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define mp make_pair #define pb push_back…