TopcoderSRM748div1-250

Problem Statement

You have a sequence S of small positive integers. You want to erase some (possibly none but not all) elements of S in such a way that the greatest common divisor of the resulting sequence becomes exactly goal.
Let W be the number of ways in which the above can be done. Compute and return the value (W modulo (10^9 + 7)).

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Definition

Class: EraseToGCD
Method: countWays
Parameters: vector , int
Returns: int
Method signature: int countWays(vector S, int goal)
(be sure your method is public)

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Limits

Time limit (s): 2.000
Memory limit (MB): 256

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Notes

Two ways of erasing are different if the sets of indices of erased elements differ.

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Constraints

S will have between 1 and 500 elements, inclusive.

Each element of S will be between 1 and 1000, inclusive.

goal will be between 1 and 1000, inclusive.

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为什么这样的套路题还可以原封不动的出出来。
设$f[i]$为答案，即$gcd$为$i$的组合有多少个，这个东西不太好求。
根据套路，要大力容斥。
设$g[i]$为$gcd$为$i$的倍数的组合的个数
那么有
$$f[i]=g[i]-\sum _{i|d且i!=d}g[d]$$
$g[i]$就比$f[i]$好求了。
设$a[i]$为数字为$i$的数字有多少个，$h[i]$为数字为$i$的倍数的数字有多少个。
那么有$$h[i]=\sum _{i|d}a[d]$$
这样就有$$g[i]=2^{h[i]}-1$$
像埃筛一样枚举，总复杂度为$O(nlogn)$

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 2000;
const int mod = 1e9+7;
typedef long long ll;
class EraseToGCD{
	public:
		ll pw(ll x,ll y){
			ll ret = 1;
			while(y){
				if(y&1)ret=(ret*x)%mod;
				x=(x*x)%mod;
				y>>=1;
			}
			return ret;
		}
		ll f[MAXN];
		int countWays(vector <int> S, int goal){
			memset(f,0,sizeof f);
			for(int i=0;i<(int)S.size();i++){
				f[S[i]]++;
			}
			for(int i=1;i<2000;i++){
				for(int j=i*2;j<2000;j+=i){
					f[i] += f[j];
				}
			}
		//	cout<<f[6]<<endl;
			for(int i=1;i<2000;i++){
				f[i]=pw(2,f[i]);
				f[i]--;
				if(f[i]<0)
					f[i]+=mod;
			}//cerr<<"here"<<endl;
			
			for(int i=1999;i>0;i--){
				for(int j=i*2;j<2000;j+=i){
					f[i] -= f[j];
					f[i]%=mod;
					if(f[i]<0)
						f[i]=(f[i]+mod)%mod;
				}
			}
			return f[goal];
		}
};

/*
int n=5,data[]={0,6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 15, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 10, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6,0},g=6;
vector<int> arr;
int main(){
	for(int i=1;data[i]!=0;i++)
		arr.push_back(data[i]);
	EraseToGCD t;
	cout<<t.countWays(arr,g);
	return 0;
}*/

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PS

当然可以反演啦。
设$f[i]$为答案，S为枚举的集合
则
f[i]=∑S[gcd{S}==i]f[i]=\sum_{S}[gcd{\{S\}} ==i]f[i]=S∑​[gcd{S}==i]
令$S‘$为$S$中的每个元素除以$i$则
f[i]=∑S‘[gcd{S‘}==1]f[i]=\sum_{S`}[gcd\{S`\}==1]f[i]=S‘∑​[gcd{S‘}==1]
然后大力反演
f[i]=∑S‘∑d∣gcd{S‘}μ(d)f[i]=\sum _{S`}\sum_{d|gcd{\{S`\}}}\mu(d)f[i]=S‘∑​d∣gcd{S‘}∑​μ(d)
参考上面的$g[i]$的定义
则
$$f[i]=\sum _d\mu (d)\ast g(i\ast d)$$
变换一下枚举元素
$$f[i]=\sum _{i|j}\mu (j/i)\ast g(j)$$
和上面一样求$g[i]$就好了。
然而太懒了，没写代码。