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Sky Code
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 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. 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 Source |
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给出一个序列求gcd为1的四元组的个数，容斥搞一下。
#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
#include <queue>
#define foreach(it,v) for(__typeof(v.begin()) it = v.begin(); it != v.end(); ++it)
using namespace std;
typedef long long ll;
const int maxn = 1e4 + 5;
bool check[maxn];
vector<int> v[maxn];
int g[maxn],a[maxn];//g[i] 表示i的倍数有多少个（只考虑素因数分解式中素数幂不超过1的i,其余为0)
void init(int n)
{
memset(check, 0, sizeof check);
for(int i = 2; i <= n; i++) {
if(check[i])continue;
for(int j = i; j <= n; j += i)
v[j].push_back(i),check[j] = 1;
}
}
void Modify(int x,int d)
{
vector<int> &cur = v[x];
int M,sz = cur.size();
M = 1<<sz;
for(int s = 0; s < M; s++) {
int now = 1;
for(int j = 0; j < sz; j++)if((s>>j)&1){
now *= cur[j];
}
g[now] += d;
}
}
ll gao(int x)
{
ll t = g[x];
if(t < 4)return 0;
t = (t*(t-1)*(t-2)*(t-3))/24;
if(v[x].size()&1) t = -t;
return t;
}
int main(int argc, char const *argv[])
{
init(maxn-5);
int n;
while(~scanf("%d",&n)) {
memset(g,0,sizeof g);
int M = 0;
for(int i = 1; i <= n; i++) {
scanf("%d",a + i);
M = max(M,a[i]);
Modify(a[i],1);
}
ll ans = 0;
for(int i = 1; i <= M; i++) ans += gao(i);
printf("%I64d\n", ans);
}
return 0;
}