P6825 「EZEC-4」求和
P6825 「EZEC-4」求和
求:
先化简原式,有:
再看后面那部分,令 。
设 为等比数列前 项和,则有:
即可 求解。
时间复杂度 。
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pll pair<ll, ll>
#define mpr make_pair
#define fi first
#define se second
const int _ = 1.5e6 + 10;
int mod;
int read()
{
int x = 0, f = 1;
char c = getchar();
while (c < '0' || c > '9')
{
if (c == '-')
f = -1;
c = getchar();
}
while (c >= '0' && c <= '9')
{
x = x * 10 + c - '0';
c = getchar();
}
return x * f;
}
int qpow(int x, int y)
{
int res = 1;
while (y)
{
if (y & 1)
res = 1ll * res * x % mod;
x = 1ll * x * x % mod;
y >>= 1;
}
return res;
}
pll qsum(int a, int n)
{
if (n == 0)
return mpr(1, 1);
if (n == 1)
return mpr(a, a);
pll tmp = qsum(a, n / 2);
if (n & 1)
return mpr((tmp.fi * (tmp.se + 1) % mod + tmp.se * tmp.se % mod * a % mod) % mod, tmp.se * tmp.se % mod * a % mod);
else
return mpr(tmp.fi * (tmp.se + 1) % mod, tmp.se * tmp.se % mod);
}
int T, n, m, k;
bitset<_> vis;
int cnt, pri[_], mu[_];
void init(int n)
{
mu[1] = 1;
for (int i = 2; i <= n; i++)
{
if (!vis[i])
pri[++cnt] = i, mu[i] = -1;
for (int j = 1; j <= cnt && i * pri[j] <= n; j++)
{
vis[i * pri[j]] = 1;
if (i % pri[j] == 0)
break;
mu[i * pri[j]] = -mu[i];
}
}
}
int solve(int n)
{
int ans = 0;
for (int d = 1; d <= n; d++)
{
int a = qpow(d, d);
int b = a;
for (int p = 1; p <= n / d; p++)
{
pll tmp = qsum(b, n / (d * p));
ans = (ans + tmp.fi * tmp.fi % mod * (mu[p] + mod) % mod) % mod;
b = 1ll * b * a % mod;
}
}
return ans;
}
signed main()
{
T = read();
init(_ - 10);
while (T--)
{
n = read(), mod = read();
printf("%d\n", solve(n));
}
return 0;
}
本文来自博客园,作者:蒟蒻orz,转载请注明原文链接:https://www.cnblogs.com/orzz/p/18122054