P1447 [NOI2010]能量采集

式子化簡

顯然題目就是要我們求${\sum }_{i=1}^n{\sum }_{j=1}^{m}2gcd(i,j)?1$

$$=2\sum _{i=1}^n\sum _{j=1}^{m}gcd(i,j)?nm$$

轉化為我們要求${\sum }_{i=1}^n{\sum }_{j=1}^{m}gcd(i,j)$

$$=\sum _{d=1}^{n}d\sum _{i=1}^{\frac{n}{d}}\sum _{j=1}^{\frac{m}{d}}gcd(i,j)==1$$

套上$mobius$

$$=\sum _{d=1}^{n}d\sum _{i=1}^{\frac{n}{d}}\sum _{j=1}^{\frac{m}{d}}\sum _{k\mid gcd(i,j)}\mu (k)$$

$$=\sum _{d=1}^{n}d\sum _{i=1}^{\frac{n}{d}}\sum _{j=1}^{\frac{m}{d}}\sum _{k\mid gcd(i,j)}\mu (k)$$

$$=\sum _{d=1}^{n}d\sum _{k=1}^{\frac{n}{d}}\mu (k)?\frac{n}{dk}??\frac{m}{dk}?$$

另$t=dk$

$$=\sum _{t=1}^n?\frac{n}{t}??\frac{m}{t}?\sum _{d\mid t}d\mu (\frac{t}{d})$$

$mobius$反演有${\sum }_{d\mid n}\frac{\mu (d)}{d}=\frac{?(n)}{n}$

$$=\sum _{t=1}^n?\frac{n}{t}??\frac{m}{t}??(t)$$

代碼

/*Author : lifehappy */ #pragma GCC optimize(2) #pragma GCC optimize(3) #include <bits/stdc++.h>#define mp make_pair #define pb push_back #define endl '\n' #define mid (l + r >> 1) #define lson rt << 1, l, mid #define rson rt << 1 | 1, mid + 1, r #define ls rt << 1 #define rs rt << 1 | 1using namespace std;typedef long long ll; typedef unsigned long long ull; typedef pair<int, int> pii;const double pi = acos(-1.0); const double eps = 1e-7; const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x; }const int N = 1e7 + 10;bool st[N];vector<int> prime;int n, m;ll phi[N];void mobius() {st[0] = st[1] = phi[1] = 1;for(int i = 2; i < N; i++) {if(!st[i]) {prime.pb(i);phi[i] = i - 1;}for(int j = 0; j < prime.size() && i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) {phi[i * prime[j]] = phi[i] * prime[j];break;}phi[i * prime[j]] = phi[i] * (prime[j] - 1);}}for(int i = 1; i < N; i++) phi[i] += phi[i - 1]; }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);mobius();ll n = read(), m = read();if(n > m) swap(n, m);ll ans = 0;for(ll l = 1, r; l <= n; l = r + 1) {r = min(n / (n / l), m / (m / l));ans += (n / l) * (m / l) * (phi[r] - phi[l - 1]);}printf("%lld\n", 2 * ans - n * m);return 0; }

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