NJU 1017 [JSCPC2016]Heresy 莫比烏斯反演

題目鏈接

題意

求

∑i=1n∑j=1mi2j2gcd(i,j)

（題面上寫 n≤106 ，事實上是 n≤1e6 .
十分感謝WuBaizhe，不然我就一直RE死不瞑目了…
莫比烏斯反演這塊也是一篇一篇看着WuBaizhe的blog學的，對初學者十分友好，每篇的推導都很詳細，非常感謝原Po）

初級版(280ms)

推導

令 gcd(i,j)=k ，則 gcd(ik,jk)=1 ，則

原式=∑i=1n∑j=1mi2j2k[gcd(ik,jk)=1]=∑k=1min(n,m)k∑i=1⌊nk⌋∑j=1⌊mk⌋(ik)2(jk)2[gcd(i,j)=1]=∑k=1min(n,m)k5∑d=1min(⌊nk⌋,⌊mk⌋)μ(d)∑i=1⌊nkd⌋∑j=1⌊mkd⌋(id)2(jd)2=∑k=1min(n,m)k5∑d=1min(⌊nk⌋,⌊mk⌋)d4μ(d)∑i=1⌊nkd⌋i2∑j=1⌊mkd⌋j2

O(n) 線性篩出 d4μ(d) ，算出 k5 與 i2 的前綴和，再兩次分塊 n√∗n√=n 就能搞定。
預處理 O(n) ，每組數據 O(n) 。

Code

#include <cstdio>
#include <iostream>
#define maxn 1000010
#define maxm maxn + 10
using namespace std;
typedef long long LL;
const LL mod = 1e9 + 7;
bool check[maxm];
int prime[maxm], kas;
LL h[maxm], pre[maxm], sum5[maxm], sum2[maxm];
LL poww(LL a, LL b) {
    LL ret = 1;
    while (b) {
        if (b & 1) ret = ret * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ret;
}
void init() {
    h[1] = 1; int tot = 0;
    for (int i = 2; i <= maxn; ++i) {
        if (!check[i]) {
            prime[tot++] = i;
            h[i] = (-poww(i, 4) + mod) % mod;
        }
        for (int j = 0; j < tot; ++j) {
            if (i * prime[j] > maxn) break;
            check[i * prime[j]] = true;
            if (i % prime[j] == 0) {
                h[i * prime[j]] = 0;
                break;
            }
            h[i * prime[j]] = (-h[i] * poww(prime[j], 4) % mod + mod) % mod;
        }
    }
    for (int i = 1; i <= maxn; ++i) {
        pre[i] = (pre[i - 1] + h[i]) % mod;
        sum5[i] = (sum5[i - 1] + poww(i, 5) % mod) % mod;
        sum2[i] = (sum2[i - 1] + poww(i, 2) % mod) % mod;
    }
}
LL calc(int a, int b) {
    int lim = min(a, b), le, ri;
    LL ret = 0;
    for (int i = 1; i <= lim; i = ri + 1) {
        le = i, ri = min(a / (a / i), b / (b / i));
        ret = (ret + (pre[ri] - pre[le - 1] + mod) * sum2[a / i] % mod * sum2[b / i] % mod + mod) % mod;
    }
    return ret;
}
void work() {
    int n, m;
    scanf("%d%d", &n, &m);
    int lim = min(n, m), le, ri;
    LL ans = 0;
    for (int i = 1; i <= lim; i = ri + 1) {
        le = i, ri = min(n / (n / i), m / (m / i));
        ans = (ans + (sum5[ri] - sum5[le - 1] + mod) % mod * calc(n / i, m / i) % mod + mod) % mod;
    }
    printf("Case #%d: %lld\n", ++kas, ans);
}
int main() {
    init();
    int T;
    scanf("%d", &T);
    while (T--) work();
    return 0;
}

升級版(96ms)

推導

對上面的式子繼續推導，令 T=kd ，

上式=∑k=1min(n,m)k5∑d=1min(⌊nk⌋,⌊mk⌋)d4μ(d)∑i=1⌊nkd⌋i2∑j=1⌊mkd⌋j2=∑T=1min(n,m)∑i=1⌊nT⌋i2∑j=1⌊mT⌋j2∑d|T(Td)5d4μ(d)=∑T=1min(n,m)∑d|TT5dμ(d)∑i=1⌊nT⌋i2∑j=1⌊mT⌋j2

令 f(T)=∑d|TT5dμ(d) ，顯然也是可以 O(n) 線性篩出的，接下來再一次分塊就行。
預處理 O(n) ，每組數據 O(n√) 。

Code

#include <cstdio>
#include <iostream>
#define maxn 1000010
#define maxm maxn + 10
using namespace std;
typedef long long LL;
const LL mod = 1e9 + 7;
bool check[maxm];
int prime[maxm], kas;
LL h[maxm], pre[maxm], sum5[maxm], sum2[maxm];
LL poww(LL a, LL b) {
    LL ret = 1;
    while (b) {
        if (b & 1) ret = ret * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ret;
}
void init() {
    h[1] = 1; int tot = 0;
    for (int i = 2; i <= maxn; ++i) {
        if (!check[i]) {
            prime[tot++] = i;
            h[i] = (poww(i, 5) - poww(i, 4) + mod) % mod;
        }
        for (int j = 0; j < tot; ++j) {
            if (i * prime[j] > maxn) break;
            check[i * prime[j]] = true;
            if (i % prime[j] == 0) {
                h[i * prime[j]] = h[i] * poww(prime[j], 5) % mod;
                break;
            }
            h[i * prime[j]] = h[i] * h[prime[j]] % mod;
        }
    }
    for (int i = 1; i <= maxn; ++i) {
        pre[i] = (pre[i - 1] + h[i]) % mod;
        sum2[i] = (sum2[i - 1] + poww(i, 2) % mod) % mod;
    }
}
void work() {
    int n, m;
    scanf("%d%d", &n, &m);
    int lim = min(n, m), le, ri;
    LL ans = 0;
    for (int i = 1; i <= lim; i = ri + 1) {
        le = i, ri = min(n / (n / i), m / (m / i));
        ans = (ans + (pre[ri] - pre[le - 1] + mod) % mod * sum2[n / i] % mod * sum2[m / i] % mod) % mod;
    }
    printf("Case #%d: %lld\n", ++kas, ans);
}
int main() {
    init();
    int T;
    scanf("%d", &T);
    while (T--) work();
    return 0;
}