1161 - Extreme GCD

1161 - Extreme GCD

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Time Limit: 1 second(s)

Memory Limit: 32 MB

All of you know that GCD means the greatest common divisor. So, you must have thought that this problem requires finding some sort of GCD. Don't worry, you are absolutely right!

Given N positive integers, not necessarily distinct, how many ways you can take 4 integers from the N numbers such that their GCD is 1.

Input

Input starts with an integer T (≤ 20), denoting the number of test cases.

Each case starts with an integer N (4 ≤ N ≤ 10000). The next line contains N integers separated by spaces. The integers will be positive and not greater than 10000.

Output

For each case, print the case number and the number of ways you can take the integers as mentioned above.

Sample Input	Output for Sample Input
3 4 2 4 6 1 5 1 2 4 6 8 10 12 46 100 131 5 6 7 8 9 10	Case 1: 1 Case 2: 4 Case 3: 195

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SPECIAL THANKS: JANE ALAM JAN (SOLUTION, DATASET)

思路：容斥原理;

http://www.cnblogs.com/zzuli2sjy/p/5467008.html和这个一样；

  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<iostream>
  4 #include<string.h>
  5 #include<queue>
  6 #include<stack>
  7 #include<map>
  8 #include<math.h>
  9 using namespace std;
 10 typedef long long LL;
 11 bool prime[10005]= {0};
 12 int ans[10005];
 13 int aa[10005];
 14 int bt[10005];
 15 int cc[10005]= {0};
 16 bool dd[10005]= {0};
 17 queue<int>que;
 18 int main(void)
 19 {
 20         int i,j,k;
 21         for(i=2; i<200; i++)
 22         {
 23                 for(j=i; i*j<=10000; j++)
 24                 {
 25                         prime[i*j]=true;
 26                 }
 27         }
 28         int cnt=0;
 29         for(i=2; i<=10000; i++)
 30         {
 31                 if(!prime[i])
 32                 {
 33                         ans[cnt++]=i;
 34                 }
 35         }int d;
 36         cin>>d;int s;
 37         for(s=1;s<=d;s++)
 38         {       cin>>k;
 39                 memset(bt,0,sizeof(bt));
 40                 for(i=0; i<k; i++)
 41                 {
 42                         scanf("%d",&aa[i]);
 43                 }
 44                 for(i=0; i<k; i++)
 45                 {
 46                         int nn=aa[i];
 47                         int t=0;
 48                         int flag=0;
 49                         while(nn>1)
 50                         {
 51                                 if(flag==0&&nn%ans[t]==0)
 52                                 {
 53                                         flag=1;
 54                                         que.push(ans[t]);
 55                                         nn/=ans[t];
 56                                 }
 57                                 else if(nn%ans[t]==0)
 58                                 {
 59                                         nn/=ans[t];
 60                                         flag=1;
 61                                 }
 62                                 else
 63                                 {
 64                                         flag=0;
 65                                         t++;
 66                                 }
 67                         }
 68                         if(nn>1)
 69                         {
 70                                 que.push(nn);
 71                         }
 72                         int xx=0;
 73                         while(!que.empty())
 74                         {
 75                                 cc[xx++]=que.front();
 76                                 que.pop();
 77                         }
 78                         int x;
 79                         int y;
 80                         for(x=1; x<=(1<<xx)-1; x++)
 81                         {
 82                                 int ak=1;
 83                                 int vv=0;
 84                                 for(j=0; j<xx; j++)
 85                                 {
 86                                         if(x&(1<<j))
 87                                         {
 88                                                 vv++;
 89                                                 ak*=cc[j];
 90                                         }
 91                                 }
 92                                 bt[ak]+=1;
 93                                 if(vv%2)
 94                                         dd[ak]=true;
 95                         }
 96                 }
 97                 LL sum=0;
 98                 LL sum1=0;
 99                 for(i=2; i<=10000; i++)
100                 {
101                         if(bt[i]>=4)
102                         {
103                                 LL nn=(LL)bt[i]*(LL)(bt[i]-1)*(LL)(bt[i]-2)*(LL)(bt[i]-3)/24;
104                                 if(dd[i])
105                                         sum+=nn;
106                                 else sum-=nn;
107                         }
108                 }
109                 sum1=(LL)k*(LL)(k-1)*(LL)(k-2)*(LL)(k-3)/24;
110                 sum1-=sum;printf("Case %d: ",s);
111                 printf("%lld\n",sum1);
112         }
113         return 0;
114 }