Baom and Fibonacci 整除分块 + 杜教筛
题意
定义(fib_i)为斐波那契数列
计算
[f(n) = sum_{i=1}^nsum_{j=1}^ngcd(fib(i),fib(j)) (mod 998244353)
]
[1leq 10^{10}
]
分析
[f(n) = sum sum fib_{(i,j)} \
= sum_{d=1}^n fib(d)sum sum [(i,j)=d]\
= sum_{d=1}^n fib(d)sum_{i=1}^{lfloorfrac{n}{d}
floor} sum_{j=1}^{lfloorfrac{n}{d}
floor}[(i,j) = 1]\
= sum_{d=1}^n fib(d)(2 sum phi(lfloorfrac{n}{d}
floor) - 1)
]
于是矩阵快速幂计算fib + 杜教筛计算(sum phi) 即可
复杂度正确性在于杜教筛记忆化搜索
代码
struct mat{
ll a[2][2];
};
mat mat_mul(mat x,mat y){
mat res;
memset(res.a,0,sizeof(res.a));
for(int i=0;i<2;i++)
for(int j=0;j<2;j++)
for(int k=0;k<2;k++)
res.a[i][j]=(res.a[i][j]+x.a[i][k]*y.a[k][j] % MOD)%MOD;
return res;
}
ll mat_pow(ll n){
mat c,res;
c.a[0][0]=c.a[0][1]=c.a[1][0]=1;
c.a[1][1]=0;
memset(res.a,0,sizeof(res.a));
for(int i=0;i<2;i++) res.a[i][i]=1;
while(n)
{
if(n&1) res=mat_mul(res,c);
c=mat_mul(c,c);
n >>= 1;
}
return res.a[0][1];
}
const int maxx=4700000;
const int mod=998244353;
map<ll,ll> P;
bool isP[maxx];
int prime[maxx];
int cnt;
ll phi[maxx];
ll inv=499122177;
void init(){
phi[1]=1;
for(int i=2;i<maxx;i++)
{
if(!isP[i]){prime[cnt++]=i;phi[i]=i-1;}
for(int j=0;j<cnt&&(ll)i*prime[j]<maxx;j++)
{
isP[i*prime[j]]=true;
if(i%prime[j])
phi[i*prime[j]]=phi[i]*(prime[j]-1);
else
{
phi[i*prime[j]]=phi[i]*prime[j];
break;
}
}
}
for(int i=1;i<maxx;i++)
phi[i]=(phi[i]+phi[i-1])%mod;
}
ll work(ll x){
if(x<maxx)return phi[x];
if(P[x])return P[x];
ll ans=((x+1)%mod*(x%mod)%mod)*inv%mod;
for(ll i=2,last;i<=x;i=last+1)
{
last=x/(x/i);
ans=(ans-(last-i+1)*work(x/i)%mod)%mod;
}
ans=(ans+mod)%mod;
P[x]=ans;
return ans;
}
ll getf(ll i,ll j){
return (mat_pow(j + 2) - mat_pow(i + 1) + MOD) % MOD;
}
int main(){
init();
ll n = rd();
ll ans = 0;
for(ll i = 1,j;i <= n;i = j + 1) {
j = n / (n / i);
ans += (2 * work(n / i) % MOD - 1 + MOD) % MOD * getf(i,j) % MOD;
ans %= MOD;
//cout << i << ' ' << j << ' ' << ans << '
';
}
printf("%lld",ans);
}