B - Sky Code Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 3904 Description Stancu likes space travels but he is a poor software developer and will never be able to buy his own spacecraft. Tha…

N. Sky Code Time Limit: 1000ms Case Time Limit: 1000ms Memory Limit: 65536KB   64-bit integer IO format: %lld      Java class name: Main Submit Status Font Size: + - Stancu likes space travels but he is a poor software developer and will never be abl…

题意 Language:Default Sky Code Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 3980 Accepted: 1333 Description Stancu likes space travels but he is a poor software developer and will never be able to buy his own spacecraft. That is why he is…

POJ3094 Sky Code(莫比乌斯反演) Sky Code 题意 给你\(n\le 10^5\)个数,这些数\(\le 10^5\),问这些这些数组成的互不相同的无序四元组(a,b,c,d)使得gcd(a,b,c,d)=1的四元组有多少? 解法 枚举一个约数\(k\),看看总共有多少个数\(S_k=\{x\}\)满足\(k|x\).那么可以保证(a,b,c,d)有的一个共同的因子是k,这样的四元组的个数就是 \[ F(k)={|S_k|\choose 4} \] 这样算会算重,比如枚举到…

题目链接:http://poj.org/problem?id=3904 Sky Code Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2968   Accepted: 998 Description Stancu likes space travels but he is a poor software developer and will never be able to buy his own spacecraft…

Sky Code 给出n个数,求选出4个数组合,使其gcd为1,,\(n<=10000\),每个数\(<=10000\). 解 理解1:容斥原理 注意到Mobius反演式子不好写出,于是我们考虑它的兄弟,容斥,于是设\(F(d)\)表示数中有约数d的个数,所以由容斥原理,我们不难得到 \[ans=\sum_{d=1}^{10000}F(d)\mu(d)\] 预处理出函数\(\mu\),和\(F\),代入式子枚举即可. 理解2:Mobius反演 考虑到无法写出具体的式子,于是我们可以列出抽象式子…

Sky Code Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2085   Accepted: 665 Description Stancu likes space travels but he is a poor software developer and will never be able to buy his own spacecraft. That is why he is preparing to ste…

题意:给定n个数ai, ai <= 10000, n <= 10000, 从中选出4个数要求gcd为1,这样的集合有多少个? 分析:首先总共集合nCr(n, 4) = n*(n-1)*(n-2)*(n-3)/24个,那么需要减掉gcd >=2 的集合.先减掉gcd为各个素数的方案数目,然后再由这些素数组成一些因子,考虑gcd为这些因子的情况.最后总结起来就是,素数个数为奇数时,减去:素数个数为偶数时,加上.具体实现的时候只要对每个ai分解质因数,然后单独考虑他的素因子能组成哪些数,这样再…

http://poj.org/problem?id=3904   题意:给一些数,求在这些数中找出四个数互质的方案数.   莫比乌斯反演的式子有两种形式http://blog.csdn.net/outer_form/article/details/50588307 这里用的是第二种形式. 求出四个数的公约数为x的倍数的方案数,即可得到,四个数的公约数为x的方案数. 这里x为1. 代码 #include<cstdio> #include<cstring> #include<io…

[题目链接] http://poj.org/problem?id=3904 [算法] 问题可以转化为求总的四元组个数 - 公约数不为1的四元组个数 总的四元组个数为C(n,4),公约数不为1的四元组个数可以用容斥原理求 [代码] #include <algorithm> #include <bitset> #include <cctype> #include <cerrno> #include <clocale> #include <cma…