反演一下可以得到$b_i=\sum\limits_{d=1}^i{\mu(i)(\lfloor \frac{i}{d} \rfloor})^n$
整除分块的话会T, 可以维护一个差分, 优化到$O(nlogn+klogk)$
#include <iostream> #include <algorithm> #include <cstdio> #include <math.h> #include <set> #include <map> #include <queue> #include <string> #include <string.h> #include <bitset> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ‘\n‘ #define DB(a) ({REP(i,1,n) cout<<a[i]<<‘ ‘;hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} //head const int N = 2e6+10; int n, k, mu[N]; ll po[N], sum[N]; int main() { scanf("%d%d", &n, &k); mu[1] = 1; REP(i,1,N-1) { for (int j=2*i; j<=N-1; j+=i) mu[j]-=mu[i]; po[i] = qpow(i,n); } ll ans = 0; REP(i,1,k) { if (mu[i]) for (int j=i; j<=k; j+=i) { sum[j]+=mu[i]*(po[j/i]-po[j/i-1]); } (sum[i]+=sum[i-1])%=P; if (sum[i]<0) sum[i]+=P; ans+=i^sum[i]; } printf("%lld\n", (ans%P+P)%P); }
Coprime Arrays CodeForces - 915G (数论水题)
原文:https://www.cnblogs.com/uid001/p/10639078.html