題面

Solution1:

\[ \begin{aligned} &\sum_{i=1}^n\sum_{i=1}^nijgcd(i,j) \\ =&\sum_{d=1}^dd\sum_{i=1}^{\frac{n}vt6mr5x}\sum_{j=1}^{\frac{n}vt6mr5x}ijd^2[\ gcd(i, j)=1\ ]\\ =&\sum_{d=1}^nd^3\sum_{i=1}^{\frac{n}vt6mr5x}\sum_{j=1}^{\frac{n}vt6mr5x}ij\sum_{k|(i,j)}\mu(k)\\ =&\sum_{d=1}^{n}d^3\sum_{k=1}^{\frac{n}vt6mr5x}\mu(k)\sum_{i=1}^{\frac{n}{dk}}\sum_{i=1}^{\frac{n}{dk}}ijk^2\\ =&\sum_{d=1}^{n}d^3\sum_{k=1}^{\frac{n}vt6mr5x}\mu(k)k^2(1+2+3+...+\frac{n}{dk})^2\\ =&\sum_{T=1}^{n}(\frac{(1+T)T}{2})^2\sum_{d|T}d^3(\frac{T}vt6mr5x)^2\mu(\frac{T}vt6mr5x)\ \ \ \ \ \ \ (ps:T=dk)\\ =&\sum_{T=1}^{n}(\frac{(1+T)T}{2})^2T^2\sum_{d|T}d\mu(\frac{T}vt6mr5x)\\ =&\sum_{T=1}^{n}(\frac{(1+T)T}{2})^2T^2\varphi(T)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (ps:Id* \mu=\varphi) \end{aligned} \]

對(duì)于前面的整數(shù)分塊，后面的考慮用杜教篩快速篩出來(lái)。

設(shè) \(f(n)=n^2\varphi(n)\),

\(g(n)=n^2\),

\(S(n)=\sum_{i=1}^nf(i)\)

\[ \begin{aligned} &g(1)S(n)\\ &=\sum_{i=1}^n(f*g)(i) - \sum_{i=2}^ng(i)S(\frac{n}{i})\\ &=\sum_{i=1}^n\sum_{d|i}d^2\varphi(d)(\frac{i}vt6mr5x)^2-\sum_{i=2}^ng(i)S(\frac{n}{i})\\ &=\sum_{i=1}^ni^2\sum_{d|i}\varphi(d)-\sum_{i=2}^ng(i)S(\frac{n}{i})\ \ \ \ \ \ \ \ \ \ \ \ (ps:1*\varphi=Id)\\ &=\sum_{i=1}^ni^3-\sum_{i=1}^ni^2S(\frac{n}{i}) \end{aligned} \]

然后直接杜教篩.(請(qǐng)仔細(xì)思考為什么將 \(g\) 構(gòu)造成這樣)

Solution2:

\[ \begin{aligned} &\sum_{i=1}^n\sum_{j=1}^nijgcd(i,j) \\ &=\sum_{i=1}^n\sum_{j=1}^nij\sum_{d|(i,j)}\varphi(d)\\ &=\sum_{d=1}^n\varphi(d)d^2\sum_{i=1}^{\frac{n}vt6mr5x}\sum_{j=1}^{\frac{n}vt6mr5x}ij\\ &=\sum_{d=1}^n\varphi(d)d^2(\sum_{i=1}^{\frac{n}vt6mr5x}i)^2\\ &=\sum_{d=1}^n\varphi(d)d^2\sum_{i=1}^{\frac{n}vt6mr5x}i^3 \end{aligned} \]

同樣杜教篩前面的。

Source:

#include <iostream> #include <cstdio> #include <map> #include <cstring>using namespace std;#define LL long long #define int long longconst int maxN = 1e6 + 1;LL Mod, inv2, inv4, n, inv6; LL qpow(LL a, LL b, LL p) {LL res = 1;while (b) { if (b & 1) res = res * a % p; a = a * a % p; b >>= 1; }return res ; } bool vis[maxN]; int prime[348514], tot; LL f[maxN]; void Init() {f[1] = 1;for (int i = 2; i <= maxN; ++ i) {if (!vis[i]) {prime[++tot] = i;f[i] = i - 1;}for (int j = 1; j <= tot && prime[j] * i <= maxN; ++ j) {vis[prime[j] * i] = 1;if (i % prime[j] == 0) {f[i * prime[j]] = f[i] * prime[j];break;} else f[i * prime[j]] = f[i] * f[prime[j]];}}for (int i = 1; i <= maxN; ++ i) f[i] = (f[i - 1] + f[i] * i % Mod * i % Mod) % Mod; }LL Sum1(LL x) { x %= Mod; return (1 + x) * x % Mod * inv2 % Mod; } LL Sum2(LL x) { x %= Mod; return (1 + x) * x % Mod * (x + x + 1) % Mod * inv6 % Mod; } LL Sum3(LL x) { x %= Mod; return x * x % Mod * (x + 1) % Mod * (x + 1) % Mod * inv4 % Mod; }map<LL, LL> F;LL Sf(LL x) {if (x <= maxN) return f[x];if (F[x]) return F[x];LL res = 0;for (int l = 2, r; l <= x; l = r + 1) {r = x / (x / l);res = (res + (Sum2(r) - Sum2(l - 1) + Mod) % Mod * Sf(x / l)) % Mod;}return F[x] = (Sum3(x) - res + Mod) % Mod; }signed main() { #ifndef ONLINE_JUDGEfreopen("P3768.in", "r", stdin);freopen("P3768.out", "w", stdout); #endifcin >> Mod >> n;Init();inv2 = qpow(2, Mod - 2, Mod);inv4 = qpow(4, Mod - 2, Mod);inv6 = qpow(6, Mod - 2, Mod);LL ans = 0;for (int l = 1, r; l <= n; l = r + 1) {r = n / (n / l);LL a = Sum1(n / l);a = a * a % Mod;ans = (ans + a * (Sf(r) - Sf(l - 1) + Mod) % Mod) % Mod;}cout << ans << endl; }

轉(zhuǎn)載于:https://www.cnblogs.com/cnyali-Tea/p/10421115.html

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