CodeForces 567C Geometric Progression 类似dp的递推统计方案数

 1 #include<cstdio>
 2 #include<map>
 3 struct node
 4 {
 5     int b;//a[i]作为i1的种数
 6     long long c;//a[i]作为i2的种数
 7 };
 8 using namespace std;
 9 int a[200010], n, k;
10 long long k2, sum;
11 map<int,node>q;
12 map<int,node>::iterator j;
13 int main()
14 {
15     while (scanf("%d%d", &n, &k) == 2)
16     {
17         q.clear();
18         k2 = (long long)k*k;
19         node u;
20         sum = u.b = u.c = 0;
21         for (int i = 0; i < n; i++)
22         {
23             scanf("%d", &a[i]);
24             if (!(a[i] % k2)) q.insert(make_pair(a[i] / k2,u));//统计出所有的i3
25         }
26         if (q.empty()) { puts("0");continue; }
27         for (int i = 0; i < n; i++)
28         {
29             if (a[i] % k2 == 0)//a[i]作为i3
30             {
31                 j = q.find(a[i]/k2);//找i2
32                 if(j!=q.end()) sum += j->second.c;//累加到总方法数
33             }
34             if (a[i] % k == 0)//a[i]作为i2
35             {
36                 j = q.find(a[i] / k);//找i1
37                 if (j != q.end()) j->second.c += j->second.b;//累加到i2的方法数
38             }
39             j = q.find(a[i]);//a[i]作为i1找i1
40             if (j != q.end()) j->second.b++;//累加
41         }
42         printf("%lld\n", sum);
43     }
44     return 0;
45 }