题意:$sum_{d|n}f(d)=n^{2}-3n+2$,求$sum_{i=1}^{n}f(i)mod 10^{9}+7$ , $n leqslant 10^{9}$ $left( T leqslant 500 ight)$组数据,只有5组>$10^{6}$
题解:看了式子感觉像是反演,但是呢....
令$S(n)=sum_{i=1}^{n}f(i)$
那么$S(n)=sum_{i=1}^{n}sum_{d|i}f(d)=sum_{i=1}^{n}f(d)lfloorfrac{n}{d} floor=sum_{d=1}^{n}sum_{i=1}^{lfloorfrac{n}{d} floor} f(d)=sum_{d=1}^{n}S(lfloorfrac{n}{d} floor)=sum_{i=1}^{n}(i-1)*(i-2)$
所以$S(n)=frac{n*(n-1)*(n-2)}{3}-sum_{i=2}^{n}f(i)$
所以老套路,预处理$maxn^{frac{2}{3}}$的S(i),这个可以直接算,总复杂度也是$O(n^{frac{2}{3}})$
#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<map> #define MAXN 5000000 #define mod 1000000007 #define inv 333333336 #define ll long long using namespace std; inline ll read() { ll x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();} return x*f; } int f[MAXN+5]; map<int,ll> mp; ll calc(int x) { if(x<=MAXN)return f[x]; map<int,ll>::iterator it; if((it=mp.find(x))!=mp.end())return it->second; ll sum=1LL*x*(x-1)%mod*(x-2)%mod*inv%mod; int last; for(int i=2;i<=x;i=last+1) { last=x/(x/i); sum-=1LL*(last-i+1)*calc(x/i); while(sum<0)sum+=mod; } return (mp[x]=sum); } int main() { for(int i=1;i<=MAXN;i++) f[i]=1LL*(i-2)*(i-1)%mod; for(int i=1;i<=MAXN;i++) for(int j=i<<1;j<=MAXN;j+=i) f[j]=(f[j]-f[i]+mod)%mod; // for(int i=1;i<=10;i++)cout<<f[i]<<endl; for(int i=2;i<=MAXN;i++) f[i]=(f[i]+f[i-1])%mod; for(int t=read();t;t--) printf("%lld ",calc(read())); return 0; }