1188 最大公約數之和 V2

思路

用歐拉函數可以化簡式子如下

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

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

$$=\sum _{i=1}^n\sum _{d\mid i}d\sum _{j=1}^i(gcd(i,d)==d)?\frac{(n+1)(n)}{2}$$

$$=\sum _{i=1}^n\sum _{d\mid i}d?(\frac{i}{d})?\frac{(n+1)(n)}{2}$$

我們再通過類似于埃篩來求得${\sum }_{i=1}^n{\sum }_{d\mid i}d?(\frac{i}{d})$，接下來就可以直接輸出答案了。

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 x = 0, f = 1; 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 x * f; }const int N = 5e6 + 10;int phi[N], n;bool st[N];vector<int> prime;ll ans[N];void init() {st[0] = st[1] = 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]) {phi[i * prime[j]] = phi[i] * (prime[j] - 1);}else {phi[i * prime[j]] = phi[i] * prime[j];break;}}}for(int i = 1; i < N; i++) {for(int j = i; j < N; j += i) {ans[j] += 1ll * i * phi[j / i];}}for(int i = 1; i < N; i++) ans[i] += ans[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);init();int T;cin >> T;while(T--) {int n;cin >> n;cout << ans[n] - 1ll * (n + 1) * n / 2 << endl;}return 0; }

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