Triple
Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 125536/65536 K (Java/Others) Total Submission(s): 1365 Accepted Submission(s): 549
Problem Description Given many different integers, find out the number of triples (a, b, c) which satisfy a, b, c are co-primed each other or are not co-primed each other. In a triple, (a, b, c) and (b, a, c) are considered as same triple. Input The first line contains a single integer T (T <= 15), indicating the number of test cases. In each case, the first line contains one integer n (3 <= n <= 800), second line contains n different integers d (2 <= d < 105) separated with space.
Output For each test case, output an integer in one line, indicating the number of triples. Sample Input 1 6 2 3 5 7 11 13 Sample Output 20 Source 2011 Multi-University Training Contest 7 - Host by ECNU
给你n个数，对于其中的任意n个数，a,b,c 要么两两互斥，要么a,b,c两两不互斥...... 要你求出满足这一条件的组合数。 分析： 对于任意的三个数，a，b,c 我们知道有这些情况，0对互斥（即两两都不互斥），1对互斥，两对互斥，三对互斥（即两两互斥）。 其后思路与其相同代码：
1 #include<cstdio>
2 #include<cstring>
3 using namespace std;
4 const int maxn=100005;
5 int item[maxn];
6 int gcd(int a,int b)
7 {
8 if(b==0)return a;
9 return gcd(b,a%b);
10 }
11 int main()
12 {
13 int cas,n;
14 scanf("%d",&cas);
15 while(cas--)
16 {
17 scanf("%d",&n);
18 for(int i=0;i<n;i++)
19 scanf("%d",item+i);
20 int ans=0;
21 for(int i=0;i<n;i++)
22 {
23 int numa=0,numb=0;
24 for(int j=0;j<n;j++)
25 {
26 if(i!=j)
27 {
28 if(gcd(item[i],item[j])==1)numa++;
29 else numb++;
30 }
31 }
32 ans+=numa*numb;
33 }
34 printf("%d\n",(n*(n-1)*(n-2)/6)-ans/2);
35 }
36 return 0;
37 }
View Code
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