Codeforces Round #729 (Div. 2) C. Strange Function (數學)

• 題意: \(f(i)=x\),\(x\)為最小的不能整除\(i\)的數.求\(\sum^{n}_{i=1}f(i)\ mod\ 10^9+7\).

• 題解:首先,\(1,2,...,x-1|i\),即\(i\)一定是\(lcm(1,2,...,x-1)\)的倍數,我們現在來看\(f(i)\)的值,當\(f(i)=2\)時,\(i\)的最小值是\(1\),\(f(i)=3\)時,\(i\)的最小值是\(2\),\(f(i)=4\)時,\(i\)的最小值是\(6\),因為\(lcm(1,2,3)=6\).以此類推,最后發現\(f(i)=43\)時,\(i\ge10^{16}\).因為\(f(i)\)最小為\(2\),所以\(ans\)初始化為\(2*n\).然后接下來我們要計算答案,可以從小到大線性枚舉\(f(i)\)的值,比如說\(f(i)=4\),最小的\(i\)為\(6\),那么對於所有的\(i=k*6\),它都包含\((1,2,3)\),所以當前這個因子可以選,答案因子還在后面(因為線性,所以+1),那么當前貢獻就要+1,即\(ans+=n/lcm(1,2,3)\).以此類推即可得到答案.

• 代碼:

  #include <bits/stdc++.h>
  #define ll long long
  #define fi first
  #define se second
  #define pb push_back
  #define me memset
  #define rep(a,b,c) for(int a=b;a<=c;++a)
  #define per(a,b,c) for(int a=b;a>=c;--a)
  const int N = 1e6 + 10;
  const int mod = 1e9 + 7;
  const int INF = 0x3f3f3f3f;
  using namespace std;
  typedef pair<int,int> PII;
  typedef pair<ll,ll> PLL;
  ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
  ll lcm(ll a,ll b) {return a/gcd(a,b)*b;}
  
  
  int main() {
      ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
  	int _;
  	cin>>_;
  	while(_--){
  		ll n;
  		cin>>n;
  		ll ans=n*2;
  		ll cur=1;
  		for(int i=3;i<=43;++i){
  			cur=lcm(cur,i-1);
  			ans=(ans+n/cur)%mod;
  		}
  		cout<<ans<<'\n';	
  	}
  	
      return 0;
  }