POJ 3904 Sky Code（容斥）

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 steal the spacecraft of Petru. There is only one problem – Petru has locked the spacecraft with a sophisticated cryptosystem based on the ID numbers of the stars from the Milky Way Galaxy. For breaking the system Stancu has to check each subset of four stars such that the only common divisor of their numbers is 1. Nasty, isn’t it? Fortunately, Stancu has succeeded to limit the number of the interesting stars to N but, any way, the possible subsets of four stars can be too many. Help him to find their number and to decide if there is a chance to break the system.

Input

In the input file several test cases are given. For each test case on the first line the number N of interesting stars is given (1 ≤ N ≤ 10000). The second line of the test case contains the list of ID numbers of the interesting stars, separated by spaces. Each ID is a positive integer which is no greater than 10000. The input data terminate with the end of file.

Output

For each test case the program should print one line with the number of subsets with the asked property.

Sample Input

4 2 3 4 5 4 2 4 6 8 7 2 3 4 5 7 6 8

Sample Output

1 0 34

题目大意

在 n 个数中统计gcd(a,b,c,d)=1的个数。

解题思路

可以用总的情况，减去gcd(a,b,c,d)!=1的情况，总的情况即为 C4n ，关键是gcd(a,b,c,d)!=1的求法。 通过对 n 个数进行质因子分解，然后统计出所出现的素因子的总个数，假如 2 出现的次数总共为 a，那么代表在 n 个数中有 a 个数含有质因子 2 ，则有 Ca4 个组合的gcd等于2，其它数同理，则根据容斥原理便可求出所有gcd不为 1 的情况，如 2 出现次数总共为 a，3 出现次数总共为 b，6出现次数总共为 c，那么此时总的gcd!=1的情况是 Ca4+Cb4−Cc4 。

代码实现

#include <iostream> #include<cstdio> #include<cstring> using namespace std; #define maxn 10007 #define ll long long int countt[maxn],num[maxn]; int prime[maxn],n; ll a[maxn]; void init() { memset(a,0,sizeof(a)); for(ll i=4;i<maxn;i++) a[i]=i*(i-1)*(i-2)*(i-3)/24; } void solve(int m) { int t=0; for(int i=2; i*i<=m; i++) { if(m%i==0) prime[t++]=i; while(m%i==0) m/=i; } if(m!=1) prime[t++]=m; for(ll i=1; i<(1<<t); i++) { int k=1; int sum=0; for(ll j=0; j<t; j++) { if(i&(1<<j)) { k*=prime[j]; sum++; } } countt[k]++; num[k]=sum; } } int main() { init(); while(~scanf("%d",&n)) { memset(countt,0,sizeof(countt)); memset(num,0,sizeof(num)); int m; for(int i=0;i<n;i++) { scanf("%d",&m); solve(m); } ll ans=0; for(int i=0;i<maxn;i++) { if(countt[i]) { if(num[i]%2) ans+=a[countt[i]]; else ans-=a[countt[i]]; } } printf("%lld\n",a[n]-ans); } return 0; }