HDU 5608 function（莫比乌斯反演 + 杜教筛）题解

题意：

已知\(N^2-3N+2=\sum_{d|N}f(d)\)，求\(\sum_{i=1}^nf(i) \mod 1e9+7\)，\(n\leq1e9\)

思路：

杜教筛基础题？

很显然这里已经设了一个\(F(n) = \sum_{d|n}f(d)\)，那么由莫比乌斯反演可以得到\(f(n)=\sum_{d|n}\mu(d)F(\frac{n}{d})\)。

然后卷积可以看出卷一个\(I\)比较好，则：\((f*g)(n)=\sum_{i=1}^nF(i)-\sum_{i=2}^nS(\lfloor\frac{n}{i}\rfloor)\)。显然\(F(i)\)前缀可以通过平方和公式等推出来。那么直接递归做即可。

代码：

#include<map>
#include<set>
#include<cmath>
#include<cstdio>
#include<stack>
#include<ctime>
#include<vector>
#include<queue>
#include<cstring>
#include<string>
#include<sstream>
#include<iostream>
#include<algorithm>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const int maxn = 1e6 + 5;
const ll MOD = 1e9 + 7;
const ull seed = 131;
const int INF = 0x3f3f3f3f;

int mu[maxn], vis[maxn];
int prime[maxn], cnt;
ll sum[maxn];
ll F(ll x){
    return (1LL * x * x - 3 * x + 2) % MOD;
}
void getmu(int n){
    memset(vis, 0, sizeof(vis));
    cnt = 0;
    mu[1] = 1;
    for(int i = 2; i <= n; i++) {
        if(!vis[i]){
            prime[cnt++] = i;
            mu[i] = -1;
        }
        for(int j = 0; j < cnt && prime[j] * i <= n; j++){
            vis[prime[j] * i] = 1;
            if(i % prime[j] == 0){
                mu[i * prime[j]] = 0;
                break;
            }
            mu[i * prime[j]] = -mu[i];
        }
    }

    for(int i = 1; i <= n; i++){
        for(int j = i; j <= n; j += i){
            sum[j] = (sum[j] + mu[i] * F(j / i)) % MOD;
        }
        sum[i] = (sum[i] + sum[i - 1]) % MOD;
    }
}

int N = 1e6;
map<int, ll> mm;
ll ksc(ll a, ll b, ll p){
    ll t = a * b - (ll)((long double)a * b / p + 0.5) * p;
    return (t < 0)? t + p : t;
}
ll inv6 = 166666668, inv2 = 500000004;
ll solve(ll n){
    if(n <= N) return sum[n];
    if(mm.find(n) != mm.end()) return mm[n];
    ll ans = ksc(ksc(n, n + 1, MOD), 2 * n + 1, MOD) * inv6 % MOD;
    ans = (ans - ksc(n, n + 1, MOD) * inv2 % MOD * 3) % MOD;
    ans += 2LL * n;
    ans %= MOD;
    for(int l = 2, r; l <= n; l = r + 1){
        r = n / (n / l);
        ans -= 1LL * (r - l + 1) * solve(n / l);
        ans %= MOD;
    }
    return mm[n] = ans;

}
ll ppow(ll a, ll b, ll p){
    ll ret = 1;
    while(b){
        if(b & 1) ret = ret * a % p;
        a = a * a % p;
        b >>= 1;
    }
    return ret;
}
int main(){
    int T;
    getmu(N);
//    printf("* %lld\n", sum[N + 1]);
    scanf("%d", &T);
    mm.clear();
    while(T--){
        ll n;
        scanf("%lld", &n);
        printf("%lld\n", (solve(n) + MOD) % MOD);
    }
    return 0;
}