Codeforces1139D_CF1139DSteps to One (Mobius_DP)
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Problem:
Analysis:
After ACing E, I gave up D and spent the left 30 minutes chatting with Little Dino.
Let (f[n]) be the expected number of steps needed to make the greatest common divisor (gcd) become (1) when the gcd is (n) now, and (g(n,d)) be the number of (x(xin[1,m])) that (gcd(x, n)=d) . So we have:
[f[n]=1+sum_{d|n}frac{f[d]cdot g(n, d)}{m}]
To make it easy, multiply (m) to the equation:
[mf[n]=m+sum_{d|n}f[d]cdot g(n, d)]
Notice that (d) can be (n), and (g(n,n)) is (lfloorfrac{m}{n} floor), so we have:
[(m-lfloorfrac{m}{n} floor)f[n]=m+sum_{d|n,d eq n}f[d]cdot g(n, d)]
Now the problem become how to calculate (g(n,d)). According to the defination,
[ egin{aligned} g(n, d)&=sum_{i=1}^m[gcd(n, i)=d]&=sum_{i=1}^{lfloorfrac{m}{d} floor}[gcd(frac{n}{d},i)=1]&=sum_{i=1}^{lfloorfrac{m}{d} floor}epsilonleft(gcd(frac{n}{d},i) ight)\end{aligned} ]
where (epsilon(x)=egin{cases}1 (x=1)\0 mathrm{otherwise}end{cases}) .
According to the Mobius Theorem ( (mu * 1 = epsilon) ) :
[ egin{aligned} g(n,d)&=sum_{i=1}^{lfloorfrac{m}{d} floor}sum_{t|frac{n}{d},t|i}mu(t)&=sum_{t|frac{n}{d}}mu(t)cdot lfloor frac{m}{dt} floor end{aligned} ]
Let‘s return to (f[n]):
[(m-lfloorfrac{m}{n} floor)f[n]=m+sum_{d|n,d eq n}f[d]sum_{t|frac{n}{d}}mu(t)cdot lfloor frac{m}{dt} floor]
Preprocess the divisors of all integer (x(xin[1,m])) and then calculate (f[n]) as the equation above directly. Because the number of divisors of most integers is very small ( for integers not more than (100000), the maximum is (128) and the total number is about (10^6) to (2 imes 10^6)) , so it won‘t TLE.
At last, the answer is:
[ans=1+sum_{i=1}^{m}frac{f[i]}{m}]
Code:
#include <cstdio>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <vector>
using namespace std;
namespace zyt
{
typedef long long ll;
const int N = 1e5 + 10, p = 1e9 + 7;
vector<int> fac[N];
int n, f[N], pcnt, prime[N], mu[N];
bool mark[N];
void init()
{
for (int i = 1; i <= n; i++)
for (int j = 1; j * j <= i; j++)
if (i % j == 0)
{
fac[i].push_back(j);
if (j * j != i)
fac[i].push_back(i / j);
}
mu[1] = 1;
for (int i = 2; i <= n; i++)
{
if (!mark[i])
prime[pcnt++] = i, mu[i] = p - 1;
for (int j = 0; j < pcnt && (ll)i * prime[j] <= n; j++)
{
int k = i * prime[j];
mark[k] = true;
if (i % prime[j] == 0)
{
mu[k] = 0;
break;
}
else
mu[k] = p - mu[i];
}
}
}
int power(int a, int b)
{
int ans = 1;
while (b)
{
if (b & 1)
ans = (ll)ans * a % p;
a = (ll)a * a % p;
b >>= 1;
}
return ans;
}
int inv(const int a)
{
return power(a, p - 2);
}
int work()
{
scanf("%d", &n);
init();
f[1] = 0;
int ans = 0;
for (int i = 2; i <= n; i++)
{
for (int j = 0; j < fac[i].size(); j++)
{
int d = fac[i][j];
if (d == i)
continue;
int tmp = 0;
for (int k = 0, size = fac[i / d].size(); k < size; k++)
{
int t = fac[i / d][k];
tmp = (tmp + (ll)mu[t] * (n / d / t) % p) % p;
}
f[i] = (f[i] + (ll)tmp * f[d] % p) % p;
}
f[i] = (ll)(f[i] + n) * inv(n - n / i) % p;
}
for (int i = 1; i <= n; i++)
ans = (ans + f[i]) % p;
printf("%d", int(((ll)ans * inv(n) % p) + 1) % p);
return 0;
}
}
int main()
{
return zyt::work();
}
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