Codeforces 839D Winter is here 组合数公式/容斥

C(n,1)+2*C(n,2)+3*C(n,3)+……+n*C(n,n)= ((1+x)^n)求导,x=1

{4,8}中算到2的部分可以在4的部分减掉，也就是从大到小算，去掉自身倍数的贡献

#include<bits/stdc++.h>#include<stdio.h>#include<algorithm>#include<queue>#include<string.h>#include<iostream>#include<math.h>#include<set>#include<map>#include<vector>#include<iomanip>using namespace std;#define ll long long#define pb push_back#define FOR(a) for(int i=1;i<=a;i++)const int inf=0x3f3f3f3f;const int maxn=1e6+9; const int mod=1e9+7;int po[maxn];int dp[maxn];int cnt[maxn];int f[maxn];int main(){int n;scanf("%d",&n);po[0]=1;for(int i=1;i<=n;i++)po[i]=2*po[i-1]%mod;int c;FOR(n)scanf("%d",&c),cnt[c]++;int ans=0;int x;for(int i=maxn-1;i>1;i--){x=0;for(int j=i;j<=maxn;j+=i)x+=cnt[j];if(x){f[i]=1ll*x*po[x-1]%mod;for(int j=i+i;j<maxn;j+=i)f[i]=(f[i]-f[j]+mod)%mod;ans=(1ll*f[i]*i+ans)%mod;}}printf("%d\n",ans);}