P4900 食堂

推導

$$\begin{gathered} ans=\sum _{i=1}^n\sum _{j=1}^i\frac{i}{j}?\sum _{i=1}^n\sum _{j=1}^i?\frac{i}{j}? \\ 前項為\sum _{i=1}^{n}i\sum _{j=1}^{i}inv(j),兩次前綴和即可得到 \\ 后項為\sum _{i=1}^n\sum _{j=1}^i?\frac{i}{j}?=\sum _{i=1}^n\sum _{j=1}^{i}d(j),nloglogn+2n即可得到 \end{gathered}$$

本來想瞎開題，沒想到又碰到數學題了

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 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 = 1e6 + 10, mod = 998244353;ll quick_pow(ll a, int n) {ll ans = 1;while(n) {if(n & 1) ans = ans * a % mod;a = a * a % mod;n >>= 1;}return ans; }ll prime[N], inv[N], f1[N], f2[N], cnt;bool st[N];void init() {inv[1] = 1;for(int i = 2; i < N; i++) {if(!st[i]) {inv[i] = quick_pow(i, mod - 2);prime[++cnt] = i;}for(int j = 1; j <= cnt && 1ll * i * prime[j] < N; j++) {st[i * prime[j]] = 1;inv[i * prime[j]] = inv[i] * inv[prime[j]] % mod;if(i % prime[j] == 0) {break;}}}for(int i = 1; i < N; i++) {inv[i] = (inv[i - 1] + inv[i]) % mod;for(int j = i; j <= N; j += i) {f2[j]++;}}for(int i = 1; i < N; i++) {f1[i] = 1ll * i * inv[i] % mod;f2[i] = (f2[i - 1] + f2[i]) % mod;}for(int i = 1; i < N; i++) {f1[i] = (f1[i - 1] + f1[i]) % mod;f2[i] = (f2[i - 1] + f2[i]) % mod;} }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 = read();while(T--) {int A = read(), B = read();printf("%lld\n", ((f1[B] - f2[B] - f1[A - 1] + f2[A - 1]) % mod + mod) % mod);}return 0; }

總結

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