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…

Candy Distribution Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 499    Accepted Submission(s): 189 Problem Description WY has n kind of candy, number 1-N, The i-th kind of candy has ai. WY w…

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 >…

题目链接: Hdu 4465 Candy 题目描述: 有两个箱子,每个箱子有n颗糖果,抽中第一个箱子的概率为p,抽中另一个箱子的概率为1-p.每次选择一个箱子,有糖果就拿走一颗,没有就换另外一个箱子.问换箱子的时候,另外一个箱子中剩下糖果的期望值. 解题思路: 注意题目描述,其中任意一个箱子没有糖果,另一个箱子中剩下糖果个数的期望,而不是第一个箱子没有糖果.不是把其中一个箱子取空时,另一个箱子剩下糖果的期望,而是其中一个箱子取空再换另外一个箱子时,这个箱子的期望. 可以根据期望性质画出公式:an…

Candy Factory Time Limit: 2000ms Memory Limit: 32768KB This problem will be judged on HDU. Original ID: 478064-bit integer IO format: %I64d      Java class name: Main   A new candy factory opens in pku-town. The factory import M machines to produce h…

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…

Candy Sharing Game Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4448    Accepted Submission(s): 2718 Problem Description A number of students sit in a circle facing their teacher in the cente…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4465 参考博客:http://www.cnblogs.com/goagain/archive/2012/11/20/2778633.html 看他的分析足够了 下面的代码也是他写的,觉得优美就贴下来: #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #includ…

简单概率题,可以直接由剩余n个递推到剩余0个.现在考虑剩余x个概率为(1-p)的candy时,概率为C(2 * n - x, x) * pow(p, n + 1)  *pow(1 - p, n - x): 在写出x - 1的情况,就可以发现组合数可以直接递推,所以可以直接求.但是考虑到p可能很小,n可能很大,这样的话直接用pow函数会丢失精度,我们可以把double类型写成log10的形式,这样可以保存精度. #include<algorithm> #include<iostream&g…