P2714 四元组统计
考虑类似于SOSDP的方法，统计超集的大小，在这个集合内任选4个数，可以算出方案数，保证他们的 $\gcd$ 为 $x$ 的倍数，再差分回去，即为答案。
查看代码
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
| #include <cstdio>
using namespace std;
template <class Type>
void read (Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
flag |= c == '-';
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 3) + (x << 1) + c - '0';
if (flag) x = ~x + 1;
}
template <class Type, class ...Rest>
void read (Type &x, Rest &...y) { read(x); read(y...); }
template <class Type>
void write (Type x)
{
if (x < 0) putchar('-'), x = ~x + 1;
if (x > 9) write(x / 10);
putchar('0' + x % 10);
}
typedef long long LL;
const int N = 1e4 + 10;
int n;
LL f[N];
bool vis[N];
int cnt, p[N];
void init ()
{
vis[1] = true;
for (int i = 2; i < N; ++i)
{
if (!vis[i]) p[++cnt] = i;
for (int j = 1; j <= cnt && i * p[j] < N; ++j)
{
vis[i * p[j]] = true;
if (i % p[j] == 0) break;
}
}
}
int main ()
{
init();
while (~scanf("%d", &n))
{
for (int i = 1; i < N; ++i) f[i] = 0;
for (int a; n; --n) read(a), ++f[a];
for (int i = 1; i <= cnt; ++i)
for (int j = (N - 1) / p[i]; j; --j)
f[j] += f[p[i] * j];
for (int i = 1; i < N; ++i)
f[i] = f[i] * (f[i] - 1) * (f[i] - 2) * (f[i] - 3) / 24;
for (int i = 1; i <= cnt; ++i)
for (int j = 1; p[i] * j < N; ++j)
f[j] -= f[p[i] * j];
write(f[1]), puts("");
}
return 0;
}
|