HDU 1220 B - Cube - 相关文章

【HDU 1220 B - Cube】的更多相关文章

HDU 1220 B - Cube

http://acm.hdu.edu.cn/showproblem.php?pid=1220 一开始的做法是,先暴力算出一个面,就是n * n的面,能有多少对.记作face 然后从上开始算下来,最上一层,face个,然后第二层,有n * n * (up - 1)个. 递推下去,up是在第i层上面有多少个. #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #inc…

HDU 1220 Cube（数学，找规律）

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1220 Cube Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2627 Accepted Submission(s): 2064 Problem Description Cowl is good at solving math pro…

题解报告：hdu 1220 Cube

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1220 问题描述 Cowl擅长解决数学问题. 有一天,一位朋友问他这样一个问题:给你一个边长为N的立方体,它被与其侧面平行的平面切割成N * N * N个单位立方体. 两个单位立方体可能没有共同点或两个共同点或四个共同点. 你的工作是计算有多少对不超过两个公共点的单位立方体.处理到文件的结尾. 输入会有很多测试用例. 每个测试用例只会在一行中给出一个立方体的边长N. N是正整数(1 <= N <=…

Cube HDU - 1220（思维）

Cowl is good at solving math problems. One day a friend asked him such a question: You are given a cube whose edge length is N, it is cut by the planes that was paralleled to its side planes into N * N * N unit cubes. Two unit cubes may have no commo…

B - Cube HDU - 1220 （数学计数）

题意:一个边长为N的正方体,切割成N*N*N个单位正方体,问有多少对正方体之间有0个,2个公共点. 思路:因为正方体之间出现公共点的情况有0,2,4. 那么直接正面求,肯定不好求,那么先求出有4个公共顶点的对数m. 怎么求呢?通过观察,其实就是外面的一个面没有相邻小正方体,其他都相邻了,那就是总的小正方体的总面数-在表面的面数就可以了同时,求的是对数,所以最后不要忘了除以2. 公式 (N*N*N*6 - N*N*6)/3 在代码中只是把它化简了 #inclu…

hdu 1220 容斥

http://acm.hdu.edu.cn/showproblem.php?pid=1220 Cube Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2260 Accepted Submission(s): 1819 Problem Description Cowl is good at solving math problems…

HDU 5292 Pocket Cube 结论题

Pocket Cube 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5292 Description Pocket Cube is the 2×2×2 equivalent of a Rubik's Cube(3×3×3). The cube consists of 8 pieces, all corners. (from wiki) It was a Pocket Cube. Unfortunately, the Cube fell to t…