Codeforces Round #428 (Div. 2) D. Winter is here[数论II][容斥原理]
传送门:http://codeforces.com/contest/839/problem/D
Examples
input
3
3 3 1
output
12
input
4
2 3 4 6
output
39
Note
In the first sample the clans are {1}, {2}, {1, 2} so the answer will be 1·3 + 1·3 + 2·3 = 12
题解:当有n个数为x的倍数时 gcd为x对答案的贡献为$1*C_n^1+2*C_n^2+...+n*C_n^n$
避免重复,从后向前计算,用f[x]表示gcd为x时存在的贡献组数,例如计算x,则在x的情况基础上减去f[2x]、f[3x]....
二项式定理:$(x+y)^n=C_n^0x^ny^0+C_n^1*x^{n-1}*y^1......$令x=y=1得$0*C_n^0+1*C_n^1+2*C_n^2+...+n*C_n^n=n*2^{n-1}$
1 #define _CRT_SECURE_NO_DEPRECATE 2 #pragma comment(linker, "/STACK:102400000,102400000") 3 #include<iostream> 4 #include<cstdio> 5 #include<fstream> 6 #include<iomanip> 7 #include<algorithm> 8 #include<cmath> 9 #include<deque> 10 #include<vector> 11 #include<bitset> 12 #include<queue> 13 #include<string> 14 #include<cstring> 15 #include<map> 16 #include<stack> 17 #include<set> 18 #include<functional> 19 #define pii pair<int, int> 20 #define mod 1000000007 21 #define mp make_pair 22 #define pi acos(-1) 23 #define eps 0.00000001 24 #define mst(a,i) memset(a,i,sizeof(a)) 25 #define all(n) n.begin(),n.end() 26 #define lson(x) ((x<<1)) 27 #define rson(x) ((x<<1)|1) 28 #define inf 0x3f3f3f3f 29 typedef long long ll; 30 typedef unsigned long long ull; 31 using namespace std; 32 const int maxn = 1e6 + 5; 33 ll savepow[maxn]; 34 int cnt[maxn]; 35 ll f[maxn]; 36 int main() 37 { 38 ios::sync_with_stdio(false); 39 cin.tie(0); cout.tie(0); 40 int i, j, k, m, n; 41 cin >> n; 42 for (int i = 1; i <= n; ++i) 43 { 44 cin >> k; 45 cnt[k]++; 46 } 47 savepow[0] = 1; 48 for (int i = 1; i <= 1000000; ++i)savepow[i] = (savepow[i - 1] << 1) % mod; 49 ll ans = 0; 50 for (ll i = 1000000; i > 1; --i) 51 { 52 int sum = 0; 53 for (int j = i; j <= maxn; j += i) 54 sum += cnt[j]; 55 if (sum) 56 { 57 ll tempans = 0; 58 tempans = 1LL * sum*savepow[sum - 1] % mod; 59 for (int j = i + i; j <= maxn; j += i) 60 tempans = (tempans - f[j] + mod) % mod; 61 f[i] = tempans; 62 ans = (ans + i*f[i]) % mod; 63 } 64 } 65 cout << ans << endl; 66 return 0; 67 }