1148 - Mad Counting（数学）

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1148 - Mad Counting（数学）

1148 - Mad Counting PDF (English) Statistics Forum Time Limit: 0.5 second(s) Memory Limit: 32 MB Mob was hijacked by the mayor of the Town "TruthTown". Mayor wants Mob to count the total population of the town. Now the naive approach to this p…

lightoj 1148 Mad Counting（数学水题）

lightoj 1148 Mad Counting 链接:http://lightoj.com/volume_showproblem.php?problem=1148 题意:民意调查,每一名公民都有盟友,问最少人数. 思路:考察的知识点有两个:第一是整数相乘取上整:第二是容器大小(ps:不能算一个知识点,只能算一个坑点). 做题思路:排序,如果被调查的人有相同盟友人数(n)的个数(cnt)大于这个数+1,即:(cnt > n+1), 容器已满,只能新开辟一个容器来装盟友. 代码: #includ…

light oj 1148 - Mad Counting

1148 - Mad Counting PDF (English) Statistics Forum Time Limit: 0.5 second(s) Memory Limit: 32 MB Mob was hijacked by the mayor of the Town "TruthTown". Mayor wants Mob to count the total population of the town. Now the naive approach to this p…

LightOJ - 1148 - Mad Counting

先上题目: 1148 - Mad Counting PDF (English) Statistics Forum Time Limit: 0.5 second(s) Memory Limit: 32 MB Mob was hijacked by the mayor of the Town "TruthTown". Mayor wants Mob to count the total population of the town. Now the naive approach to…

Codeforces 911D. Inversion Counting (数学、思维)

题目链接:Inversion Counting 题意: 定义数列{ai|i=1,2,...,n}的逆序对如下:对于所有的1≤j<i≤n,若ai<aj,则<i,j>为一个逆序对.于是,对于一个数列a[1..n],给定m次操作.对于每一次操作,给定l,r(1≤l<r≤n),将序列a[l..r]倒置.求倒置后的逆序对的数量的奇偶性. 题解: 假设现在我们有一个序列并翻转这个序列[l,r]区间里面的数.假设某个数的k值是指在这个值后面小于这个数的数的个数,其实我们可以发现对于[1,l…

17997 Simple Counting 数学

17997 Simple Counting 时间限制:2000MS 内存限制:65535K提交次数:0 通过次数:0 题型: 编程题语言: 不限定 Description Ly is crazy about counting . Recently , he got a simple problem , but he had to learn Gaoshu these days .So , he turns to you for help . You are given a sequenc…

UVA 11401 - Triangle CountingTriangle Counting 数学

You are given n rods of length 1,2, . . . , n. You have to pick any 3 of them and build a triangle. Howmany distinct triangles can you make? Note that, two triangles will be considered diﬀerent if they haveat least 1 pair of arms with diﬀerent length…