（每日一题）CF1139D Steps to One && 2021年天梯赛 L3-3 可怜的简单题（期望，莫比乌斯反演，杜教筛）

Posted 2021-05-21 繁凡さん

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2021年天梯赛 L3-3 可怜的简单题

CF1139D

Problem

Solution

数据较大，要用快速乘不然会爆 long long

Code

#include <bits/stdc++.h>
using namespace std; 
#define ll long long
const int N = 21544400 + 7;

#define minus(x, y) (1ll * x - y < 0 ? 1ll * x - y + mod : 1ll * x - y)
#define plus(x, y) (1ll * x + y >= mod ? 1ll * x + y - mod : 1ll * x + y) 

ll mod;
int mu[N];
ll smu[N];
bool vis[N];
unordered_map <ll, ll> sum_mu;
int primes[N], cnt;
ll n, m; 

ll mul(ll a, ll b) {
    if (mod <= 1000000000) return a * b % mod;
    else if (mod <= 1000000000000ll) return (((a * (b >> 20) % mod) << 20) + (a * (b & ((1 << 20) - 1)))) % mod;
    else {
        ll d = (ll)floor(a * (long double)b / mod + 0.5);
        ll res = (a * b - d * mod) % mod;
        if (res < 0) res += mod;
        return res;
    }
}

ll qpow(ll a, ll b)
{
	if(mod == 1) return 0;
	ll res = 1;
	a = a % mod;
	while(b) {
		if(b & 1) res = mul(res, a);
		a = mul(a, a);
		b >>= 1;
	}
	return res;
}

ll inv(ll x)
{
	return qpow(x, mod - 2);
}

void init(int n)
{
	mu[1] = 1, vis[1] = 1;
	for(int i = 2; i <= n; ++ i) {
		if(vis[i] == 0) {
			primes[ ++ cnt] = i;
			mu[i] = -1;
		}
		for(int j = 1; j <= cnt && i * primes[j] <= n; ++ j) {
			vis[i * primes[j]] = 1;
			if(i % primes[j] == 0) {
				mu[i * primes[j]] = 0;
				break;
			}	
			mu[i * primes[j]] -= mu[i];
		}
	}	
	for(int i = 1; i <= n; ++ i) 
		smu[i] = plus(smu[i - 1], mu[i]);
	return ;
}

inline ll g_sum(ll x)
{
	return x;
}

inline ll get_sum_mu(ll x)
{
	if(x <= N - 7) return smu[x];
	if(sum_mu.find(x) != sum_mu.end()) return sum_mu[x]; 

	ll ans = 1;
	for(ll l = 2, r; l <= x; l = r + 1) {
		r = x / (x / l); 
		ll tmp = mul((r - l + 1), get_sum_mu(x / l));
		ans = minus(ans, tmp);
	}
	ans = (ans % mod + mod) % mod;
	sum_mu[x] = ans / g_sum(1ll);
	return ans / g_sum(1ll); 
}

void solve(ll m)
{
	ll ans = 1;
	for(ll l = 1, r; l <= m; l = r + 1) {
		r = m / (m / l); 
		ll tmp = mul((get_sum_mu(r) - get_sum_mu(l - 1)), mul(m / l, inv(m - m / l)));
		ans = minus(ans, tmp);
	}  
	ans = (ans + mod) % mod;
	printf("%lld\\n", ans);
	return ;
}

signed main()
{ 
	scanf("%lld%lld", &m, &mod);
	if(m == 1) return printf("1\\n"), 0;
	init(N - 7);
	solve(m);
	return 0;
}