P3768 简单的数学题 [狄利克雷卷积,杜教筛,莫比乌斯反演]
簡單的數學題
題目連接
https://www.luogu.org/problemnew/show/P3768
題目描述
輸入一個正整數n,n≤1010n,n\le 10^{10}n,n≤1010和p,p≤1.1×109p,p \le 1.1 \times 10^9p,p≤1.1×109.且ppp為質數.
計算∑i=1n∑j=1nijgcd(i,j)\sum_{i=1}^n\sum_{j=1}^nijgcd(i,j)∑i=1n?∑j=1n?ijgcd(i,j)對ppp取模.
題解
方法一.暴力推公式
原式=∑d=1nd3∑i=1n/d∑j=1n/dij×[gcd(i,j)=1]\sum_{d=1}^nd^3\sum_{i=1}^{n/d}\sum_{j=1}^{n/d}ij\times [gcd(i,j)=1]∑d=1n?d3∑i=1n/d?∑j=1n/d?ij×[gcd(i,j)=1]
設f(x)=∑i=1n/d∑j=1n/dij×[gcd(i,j)=x]f(x)=\sum_{i=1}^{n/d}\sum_{j=1}^{n/d}ij\times [gcd(i,j)=x]f(x)=∑i=1n/d?∑j=1n/d?ij×[gcd(i,j)=x]
g(x)=∑x∣df(d)=x2∑i=1n/dx∑j=1n/dxijg(x) = \sum_{x|d}f(d)=x^2\sum_{i=1}^{n/dx}\sum_{j=1}^{n/dx}ijg(x)=∑x∣d?f(d)=x2∑i=1n/dx?∑j=1n/dx?ij
故f(x)=∑x∣pμ(p/x)g(p)=∑x∣pμ(p/x)p2∑i=1n/dp∑j=1n/dpijf(x)=\sum_{x|p}\mu(p/x)g(p)=\sum_{x|p}\mu(p/x)p^2\sum_{i=1}^{n/dp}\sum_{j=1}^{n/dp}ijf(x)=∑x∣p?μ(p/x)g(p)=∑x∣p?μ(p/x)p2∑i=1n/dp?∑j=1n/dp?ij
因此f(1)=∑p=1nμ(p)p2∑i=1n/dp∑j=1n/dpijf(1)=\sum_{p=1}^n\mu(p)p^2\sum_{i=1}^{n/dp}\sum_{j=1}^{n/dp}ijf(1)=∑p=1n?μ(p)p2∑i=1n/dp?∑j=1n/dp?ij
原式=∑d=1nd3∑p=1nμ(p)p2∑i=1n/dp∑j=1n/dpij\sum_{d=1}^nd^3\sum_{p=1}^n\mu(p)p^2\sum_{i=1}^{n/dp}\sum_{j=1}^{n/dp}ij∑d=1n?d3∑p=1n?μ(p)p2∑i=1n/dp?∑j=1n/dp?ij
轉而枚舉a=dpa=dpa=dp.我們得到:
原式=∑a=1n∑d∣aa2dμ(a/d)∑i=1n/ai∑j=1n/aj=\sum_{a=1}^n\sum_{d|a}a^2d\mu(a/d)\sum_{i=1}^{n/a}i\sum_{j=1}^{n/a}j=∑a=1n?∑d∣a?a2dμ(a/d)∑i=1n/a?i∑j=1n/a?j
=∑a=1n(?na?(?na?+1)2)2a2∑d∣adμ(a/d)=\sum_{a=1}^n (\frac{\lfloor \frac{n}{a} \rfloor(\lfloor \frac{n}{a} \rfloor+1)}{2})^2a^2\sum_{d|a}d\mu(a/d)=∑a=1n?(2?an??(?an??+1)?)2a2∑d∣a?dμ(a/d)
我們發現Id?μ=?Id*\mu=\phiId?μ=?是常見的狄利克雷卷積.
因此原式=∑a=1n(?na?(?na?+1)2)2a2?(a)=\sum_{a=1}^n (\frac{\lfloor \frac{n}{a} \rfloor(\lfloor \frac{n}{a} \rfloor+1)}{2})^2a^2\phi(a)=∑a=1n?(2?an??(?an??+1)?)2a2?(a)
化簡到這一步的時候,前面部分可以分塊計算,后面的部分a2?(a)a^2\phi(a)a2?(a)要快速計算前綴和.
于是我們想到了杜教篩:
記f(x)=x2?(x)f(x)=x^2\phi(x)f(x)=x2?(x),找到積性函數g(x)=x2g(x)=x^2g(x)=x2與它做卷積使得g(x)g(x)g(x)的前綴和可以快速求出,而(f?g)(x)=∑d∣xxd2×d2?(d)=x2∑d∣x?(x)=x3(f*g)(x)=\sum_{d|x}\frac{x}ozvdkddzhkzd^2\times d^2\phi(d)=x^2\sum_{d|x}\phi(x)=x^3(f?g)(x)=∑d∣x?dx?2×d2?(d)=x2∑d∣x??(x)=x3的前綴和也可以快速求出.
記S(x)=∑i=1xf(x)S(x)=\sum_{i=1}^xf(x)S(x)=∑i=1x?f(x)
根據杜教篩公式g(1)S(n)=∑x=1n(f?g)(x)?∑x=2ng(x)S(nx)g(1)S(n)=\sum_{x=1}^n(f*g)(x) - \sum_{x=2}^ng(x)S(\frac{n}{x})g(1)S(n)=∑x=1n?(f?g)(x)?∑x=2n?g(x)S(xn?).
S(n)=∑x=1nx3?∑x=2nx2S(nx)S(n)=\sum_{x=1}^nx^3-\sum_{x=2}^nx^2S(\frac{n}{x})S(n)=∑x=1n?x3?∑x=2n?x2S(xn?)
寫到這就完了.
方法二.?\phi?卷積
根據公式:
gcd(i,j)=∑d∣gcd(i,j)?(d)=∑d∣i,d∣j?(d)gcd(i,j) = \sum_{d|gcd(i,j)}\phi(d)=\sum_{d|i,d|j}\phi(d)gcd(i,j)=∑d∣gcd(i,j)??(d)=∑d∣i,d∣j??(d)
因此
∑i=1n∑j=1nijgcd(i,j)=∑i=1n∑j=1nij∑d∣i,d∣j?(d)=∑d=1n?(d)d2∑i=1n/d∑j=1n/dij=∑d=1n?(d)d2(?nd?(?nd?+1)2)2\sum_{i=1}^n\sum_{j=1}^nijgcd(i,j)=\sum_{i=1}^n\sum_{j=1}^nij\sum_{d|i,d|j}\phi(d)=\sum_{d=1}^n\phi(d)d^2\sum_{i=1}^{n/d}\sum_{j=1}^{n/d}ij=\sum_{d=1}^n\phi(d)d^2(\frac{\lfloor \frac{n}ozvdkddzhkzd \rfloor(\lfloor \frac{n}ozvdkddzhkzd \rfloor+1)}{2})^2∑i=1n?∑j=1n?ijgcd(i,j)=∑i=1n?∑j=1n?ij∑d∣i,d∣j??(d)=∑d=1n??(d)d2∑i=1n/d?∑j=1n/d?ij=∑d=1n??(d)d2(2?dn??(?dn??+1)?)2
同樣也得到了我們的式子,是不是推導方法簡單多了?
遇見gcd(i,j)gcd(i,j)gcd(i,j)的時候,多想想是不是可以用?\phi?卷積來做?
可能會減少很多不必要的推導過程.
代碼
#include <iostream> #include <algorithm> #include <cstring> #include <unordered_map> #define pr(x) std::cout << #x << ':' << x << std::endl #define rep(i,a,b) for(LL i = a;i <= b;++i) const int N = 1e7; typedef long long LL; LL n,p; LL phi[N+10]; int prime[N+10],zhi[N+10],low[N+10],pcnt; LL mod_pow(LL x,LL n) {LL res = 1;while(n) {if(n&1) res = res * x % p;x = x * x % p;n >>= 1;}return res; } LL inv6,inv2,inv4; void sieve() {pcnt = 0;low[1] = phi[1] = zhi[1] = 1;rep(i,2,N) {if(!zhi[i]) {phi[i] = i-1;prime[pcnt++] = i;low[i] = i;}for(LL j = 0;j < pcnt && prime[j]*i <= N;++j) {zhi[i*prime[j]] = 1;if(i % prime[j] == 0) {low[i*prime[j]] = low[i] * prime[j];if(i == low[i]) {phi[i*prime[j]] = phi[i]*prime[j];}else {phi[i*prime[j]] = phi[i/low[i]] * phi[low[i]*prime[j]];}break;}else{low[i*prime[j]] = prime[j];phi[i*prime[j]] = phi[i] * phi[prime[j]];}}} } LL sum3(LL n) {n %= p;LL res = n*(n+1)%p*(2*n+1)%p*inv6%p;return res; } LL sum4(LL n) {n %= p;LL x = n*(n+1)%p;return x*x%p*inv4%p; } std::unordered_map<LL,LL> vis,rec; LL F(LL n) {if(n <= N) return phi[n];if(vis[n]) return rec[n];LL res = sum4(n);for(LL x = 2,last;x <= n;x = last) {last = n/(n/x)+1;res = (res - ((sum3(last-1)-sum3(x-1)+p)%p*F(n/x)%p) + p) % p;}vis[n] = 1;return rec[n] = res; }signed main() {sieve();std::ios::sync_with_stdio(false);std::cin >> p >> n;inv6 = mod_pow(6,p-2);inv4 = mod_pow(4,p-2);inv2 = mod_pow(2,p-2);rep(i,1,N) phi[i] = ((phi[i]*i%p*i%p) + phi[i-1]) % p;LL last,ans = 0;for(LL x = 1;x <= n;x = last+1) {last = n/(n/x);LL y = n/x%p;LL z = (1+y)*(y)/2%p;ans = (ans + (z*z%p*((F(last)-F(x-1)+p)%p))) % p;}std::cout << ans << std::endl;return 0; }總結
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