codeforces 895 C. Square Subsets(状压dp)

题目链接：http://codeforces.com/contest/895/problem/C

思路：要为平方数那就需要相同因子个数是偶数，因为ai比较小，可以先对（1-70）每个数分解质因数。然后就是一个状压dp。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <cmath>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll inff = 0x3f3f3f3f3f3f3f3f;
#define FOR(i,a,b) for(int i(a);i<=(b);++i)
#define FOL(i,a,b) for(int i(a);i>=(b);--i)
#define REW(a,b) memset(a,b,sizeof(a))
#define inf int(0x3f3f3f3f)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%I64d",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
#define mod ll(1e9+7)
#define pb push_back
#define lc (d<<1)
#define Pll pair<ll,ll>
#define P pair<int,int>
#define pi acos(-1)
int p[20]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67},n,a[78],b,dp[78][(1<<19)+5],c[78],f[100008];
int main()
{
    cin.tie(0);
    cout.tie(0);
    cin>>n;
    f[0]=1ll;
    FOR(i,1,n) sl(b),a[b]++,f[i]=(f[i-1]*2ll)%mod;
    FOR(i,1,70)
    {
        b=i;
        FOL(j,18,0)
        {
            while(b%p[j]==0)  b/=p[j],c[i]^=(1<<j);
        }
    }
    dp[0][0]=1ll;
    FOR(i,1,70)
    {
        if(!a[i])
        {
            FOR(j,0,(1<<19)) dp[i][j]=dp[i-1][j];
            continue;
        }
        FOR(j,0,(1<<19))
        {
            dp[i][j^c[i]]=(1ll*dp[i][j^c[i]]+1ll*f[a[i]-1]*1ll*dp[i-1][j])%mod;
            dp[i][j]=(1ll*dp[i][j]+1ll*f[a[i]-1]*1ll*dp[i-1][j])%mod;
        }
    }
    cout<<(dp[70][0]-1)%mod<<endl;
    return 0;
}