HDU 6053 莫比乌斯反演

那道题想到枚举1e5以内的质因子, 但是会重复

然后队友提醒了mobius反演, 果然还是太菜了啊

题解

首先, 题目提到任意区间满足条件, 也就是gcd(b[1], b[2], b[3]…b
) >= 2 就行了, 很容易得出总的b数量为

sum=∏i=1na[i]

而其中不满足条件的就是gcd(b[1], b[2], b[3]…b
) = 1的数量

我们定义

F(n)为gcd为n的倍数的b数量, f(n)为gcd为n的b数量

则有

F(n)=∏i=1a[i]/n

F(n)=∑n|df(d)

反演得:

f(n)=∑n|du(d/n)∗F(d)

所以

f(1)=∑i=1min(a[1],a[2]...a[n])u(i)∗F(i)

直接求会超时, 注意到是整数除法, 读入的时候预处理分块用快速幂可以降低时间复杂度

然后sum-f(1)即可

code:

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long ll;

const int N = 100010;
const ll mod = 1e9 + 7;

int mobius
;
int prime
;
int a
;
bool vis
;
int cnt;
int pre[N << 1];
int n;

inline ll min(ll a, ll b)
{
return a <= b ? a : b;
}
inline ll max(ll a, ll b)
{
return a >= b ? a : b;
}

ll fastPowMod(ll a, ll b, ll mod)
{
ll res = 1 % mod;
while(b)
{
if(b & 1) res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}

void getMobius()
{
cnt = 0;

memset(vis, 0, sizeof vis);
memset(mobius, 0, sizeof mobius);

mobius[1] = 1;

for(int i = 2; i <= 100000; ++i)
{
if(!vis[i])
{
prime[++cnt] = i;
mobius[i] = -1;
}

for(int j = 1; j <= cnt && i * prime[j] <= 100000; ++j)
{
vis[i * prime[j]] = 1;

if(i % prime[j] == 0)
{
mobius[i * prime[j]] = 0;
break;
}

mobius[i * prime[j]] = -mobius[i];
}
}
}

int main()
{
getMobius();
int t, kas = 0;
scanf("%d", &t);
while(t--)
{
scanf("%d", &n);
memset(pre, 0, sizeof pre);
ll mn = N, mx = 0;
ll a_multi = 1;
for(int i = 1; i <= n; ++i)
{
scanf("%d", a + i);
++pre[a[i]];
a_multi *= a[i];
a_multi %= mod;
mn = min(mn, a[i]);
mx = max(mx, a[i]);
}
for(int i = 1; i < (N << 1); ++i) pre[i] += pre[i - 1];//前缀和用于分块
ll rs = 0;
for(int i = 1; i <= mn; ++i)
{
ll r = 1;
for(ll j = 1; j <= mx / i; ++j)
{
r = r * fastPowMod(j, pre[j * i + i - 1] - pre[j * i - 1], mod) % mod;//
}
r = (r * mobius[i] % mod + mod) % mod;
rs = (rs + r) % mod;

}
printf("Case #%d: %I64d\n", ++kas, ((a_multi - rs) % mod + mod) % mod);
}

return 0;
}