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 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
思路:容斥原理；
由于给的数据范围是10000；所以我们先打表10000以内的素数;
然后我们分解每一个数；求出它的各个不同的质因数，然后暴力组合每个数的质因数，在bt数组里记录个数，也就是bt[i],i这个数可以被前面的哪些数整除
最后从1循环到10000，容斥一遍就可以得到不合要求的个数，最后总的减去就行。
由于每个数不过10000，他的质因数不会超过8个，那么复杂度为（n*28）；
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 } 36 while(scanf("%d",&k)!=EOF) 37 { 38 memset(bt,0,sizeof(bt)); 39 for(i=0; i<k; i++) 40 { 41 scanf("%d",&aa[i]); 42 } 43 for(i=0; i<k; i++) 44 { 45 int nn=aa[i]; 46 int t=0; 47 int flag=0; 48 while(nn>1) 49 { 50 if(flag==0&&nn%ans[t]==0) 51 { 52 flag=1; 53 que.push(ans[t]); 54 nn/=ans[t]; 55 } 56 else if(nn%ans[t]==0) 57 { 58 nn/=ans[t]; 59 flag=1; 60 } 61 else 62 { 63 flag=0; 64 t++; 65 } 66 } 67 if(nn>1) 68 { 69 que.push(nn); 70 } 71 int xx=0; 72 while(!que.empty()) 73 { 74 cc[xx++]=que.front(); 75 que.pop(); 76 } 77 int x; 78 int y; 79 for(x=1; x<=(1<<xx)-1; x++) 80 { 81 int ak=1; 82 int vv=0; 83 for(j=0; j<xx; j++) 84 { 85 if(x&(1<<j)) 86 { 87 vv++; 88 ak*=cc[j]; 89 } 90 } 91 bt[ak]+=1; 92 if(vv%2) 93 dd[ak]=true; 94 } 95 } 96 LL sum=0; 97 LL sum1=0; 98 for(i=2; i<=10000; i++) 99 { 100 if(bt[i]>=4) 101 { 102 LL nn=(LL)bt[i]*(LL)(bt[i]-1)*(LL)(bt[i]-2)*(LL)(bt[i]-3)/24; 103 if(dd[i]) 104 sum+=nn; 105 else sum-=nn; 106 } 107 } 108 sum1=(LL)k*(LL)(k-1)*(LL)(k-2)*(LL)(k-3)/24; 109 sum1-=sum; 110 printf("%lld ",sum1); 111 } 112 return 0; 113 }