题意
求解((sumlimits^n_{i=1}sumlimits^n_{j=1}ijgcd(i,j))mod~p)
解题过程
(sumlimits^n_{i=1}sumlimits^n_{j=1}ijgcd(i,j))
枚举gcd的值
(=sumlimits^n_{d=1}dsumlimits^n_{i=1}sumlimits^n_{j=1}ij[gcd(i,j)=d])
(=sumlimits^n_{d=1}d^3sumlimits^{lfloor frac{n}{d} floor}_{i=1}sumlimits^{lfloor frac{n}{d} floor}_{j=1}ij[gcd(i,j)=1])
先把后面的部分莫比乌斯反演一下
(=sumlimits^n_{d=1}d^3sumlimits^{lfloor frac{n}{d} floor}_{i=1}sumlimits^{lfloor frac{n}{d} floor}_{j=1}ijsumlimits_{t|gcd(i,j)}mu(t))
把(t)提前
(=sumlimits^n_{d=1}d^3sumlimits^{lfloor frac{n}{d} floor}_{t=1}mu(t)*t^2sumlimits^{lfloor frac{n}{td} floor}_{i=1}sumlimits^{lfloor frac{n}{td} floor}_{j=1}ij)
令(1+2+3+cdots+n=S(n))
那么(=sumlimits^n_{d=1}d^3sumlimits^{lfloor frac{n}{d} floor}_{t=1}mu(t)*t^2sumlimits^{lfloor frac{n}{td} floor}_{i=1}sumlimits^{lfloor frac{n}{td} floor}_{j=1}ij)相当于
(=sumlimits^n_{d=1}dsumlimits^{lfloor frac{n}{d} floor}_{t=1}mu(t)*t^2d^2S(lfloor frac{n}{td} floor)^2)
令(T=td)
(=sumlimits^n_{d=1}dsumlimits^{lfloor frac{n}{d} floor}_{t=1}mu(t)*T^2S(lfloor frac{n}{T} floor)^2)
把T提到最外面
(=sumlimits_{T=1}^nT^2S(lfloor frac{n}{T} floor)^2sumlimits_{d|T}frac{mu(d)T}{d})
由(sumlimits_{d|n}frac{mu(d)}{d}=frac{varphi(n)}{n})
得(=sumlimits_{T=1}^nT^2S(lfloor frac{n}{T} floor)^2sumlimits_{d|T}frac{mu(d)T}{d}=sumlimits_{T=1}^nT^2S(lfloor frac{n}{T} floor)^2varphi(T))
好像还是不太行
我们令(f(i)=i^2varphi(i))
注意到(sumlimits_{d|n}varphi(d)=n)
那么设一个(g(i)=i^2)
那么再看一下杜教筛的套路式子
(g(1)S(n)=sumlimits_{i=1}^{n}(f*g)(i)-sumlimits_{d=2}^{n}g(d)cdot S(lfloorfrac{n}{d} floor)=sumlimits_{i=1}^{n}i^3-sumlimits_{d=2}^{n}g(d)cdot S(lfloorfrac{n}{d} floor))
这里的(S)是(f)的前缀和,为了方别区分将原来的(1+2+3+cdots+n)称为(Sum(n))
(S(n)=sumlimits_{i=1}^{n}i^3-sumlimits_{d=2}^{n}d^2cdot S(lfloorfrac{n}{d} floor))
其实(sumlimits_{i=1}^{n}i^3=Sum(i)^2)
最后搞出来的是(sumlimits_{T=1}^nSum(lfloor frac{n}{T} floor)^2f(T))
前半部分能数论分块，后半部分能杜教筛，那么应该差不多了
代码
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/trie_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
// freopen("k.in", "r", stdin);
// freopen("k.out", "w", stdout);
// clock_t c1 = clock();
// std::cerr << "Time:" << clock() - c1 <<"ms" << std::endl;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
mt19937 rnd(time(NULL));
#define de(a) cout << #a << " = " << a << endl
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define ls ((x) << 1)
#define rs ((x) << 1 | 1)
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<char, char> PCC;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
#define inf 0x3f3f3f3f
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll MAXN = 1e7 + 7;
const ll MAXM = 4e5 + 7;
const int MOD = 1e9 + 7;
const double eps = 1e-7;
ll p, n;
gp_hash_table<ll, ll> mp;
int phi[MAXN], pri[MAXN], tot = 0;
int vis[MAXN];
ll sum[MAXN];
void init()
{
phi[1] = 1;
for (int i = 2; i < MAXN; i++)
{
if (!vis[i])
pri[++tot] = i, phi[i] = i - 1;
for (int j = 1; j <= tot && pri[j] * i < MAXN; j++)
{
vis[i * pri[j]] = 1;
if (i % pri[j] == 0)
{
phi[i * pri[j]] = phi[i] * pri[j];
break;
}
phi[i * pri[j]] = phi[i] * (pri[j] - 1);
}
}
sum[0] = 0;
for (int i = 1; i < MAXN; i++)
sum[i] = (sum[i - 1] + 1LL * phi[i] * i % p * i % p) % p;
}
void add(ll &x, ll &y) { x = x + y >= p ? x + y - p : x + y; }
void del(ll &x, ll &y) { x = x - y < 0 ? x - y + p : x - y; }
ll Inv(ll a) { return a == 1 ? 1 : (p - p / a) * Inv(p % a) % p; }
ll inv6;
ll calSum(ll x)
{
x %= p;
ll temp = x * (x + 1) / 2 % p;
temp %= p;
return temp * temp % p;
}
ll cal_sumg(ll x)
{
x %= p;
return x * (x << 1 | 1) % p * (x + 1) % p * inv6 % p;
}
ll cal_S(ll x)
{
if (x < MAXN)
return sum[x];
if (mp.find(x) != mp.end())
return mp[x];
ll ans = calSum(x);
for (ll l = 2, r = 0; l <= x; l = r + 1)
{
r = x / (x / l);
ans -= (cal_sumg(r) - cal_sumg(l - 1)) % p * cal_S(x / l) % p;
ans = ((ans % p) + p) % p;
}
return mp[x] = ans;
}
int main()
{
mp.clear();
scanf("%lld%lld", &p, &n);
init();
inv6 = Inv(6);
ll ans = 0;
for (ll l = 1, r = 0; l <= n; l = r + 1)
{
r = n / (n / l);
ll s = calSum(n / l);
ans += (cal_S(r) - cal_S(l - 1)) % p * s % p;
ans = ((ans % p) + p) % p;
}
printf("%lld
", ans);
return 0;
}