照亮数学的七道光芒 T1-4
照亮数学的七道光芒
勇气
对于第 \(k\) 次攻击，其攻击力为：
\[a_k=\frac{x^{2^k}}{2^{2^k-2}}
\]
对于这个题，显然就是找最小的 \(k\) 使满足这个式子
\[\frac{x^{2^k}}{2^{2^k-2}}\geq 2^n
\]
以 \(2\) 为底取对数，有
\[2^k\log_2x-(2^k-2) ≥ n
\]
\[2^k ≥\frac{n-2}{\log_2x -1}
\]
如果 \(x\geq 2^n\) 答案直接就是零，如果 \(x=2\) 而 \(n>2\) 则无解。
signed main()
{
cin >> x >> n;
if (x == 2 && n > 2)
{
cout << "inf";
return 0;
}
else if (x == 2 && n == 2)
{
cout << 1;
return 0;
}
else if (n == 1)
{
cout << 0;
return 0;
}
int times = (ceil)(n - 2) / (log2(x) - 1);
int check = 1;
if (check >= times)
{
cout << 0;
return 0;
}
for (rint i = 1; ; i++)
{
check *= 2;
if (check >= times)
{
cout << i;
return 0;
}
}
}
奉献
仔细读题寻找性质，当且仅当 \(\gcd(a,b)=1\) 时，它没有在之前被填写过
将朴素式子列出来：
\[\sum_{i=1}^n\sum_{j=1}^i(d_i\log_2d_i+\sum_{k=1}^{\lfloor\frac{n}{i}\rfloor}d_k)\times check(\gcd(i,j))
\]
其中 \(check(x)\) 为 bool 型，当且仅当 \(x=1\) 时为 \(1\)，否则为 \(0\)。
将式子进行转化：
\[\sum_{i=1}^n(d_i\log_2d_i+\sum_{k=1}^{\lfloor\frac{n}{i}\rfloor}d_k)\sum_{j=1}^icheck(\gcd(i,j))
\]
即为：
\[\sum_{i=1}^n(d_i\log_2d_i+\sum_{k=1}^{\lfloor\frac{n}{i}\rfloor}d_k)\times\varphi(i)
\]
这些东西我们都是会求的，这个题最后一个地方，在于 \(d_i\) 不能直接求，差分一下。
void init()
{
v[1] = phi[1] = 1;
for (rint i = 2; i <= n; i++)
{
if (!v[i])
{
prime[++cnt] = i;
phi[i] = i - 1;
}
for (rint j = 1; prime[j] * i <= n && j <= cnt; j++)
{
v[i * prime[j]] = 1;
if (!(i % prime[j]))
{
phi[i * prime[j]] = phi[i] * prime[j];
break;
}
phi[i * prime[j]] = phi[i] * phi[prime[j]];
}
}
}
signed main()
{
cin >> n;
init();
for (rint i = 1; i <= n; i++)
{
d[i] = 1 + d[i / 10];
s[i] = s[i - 1] + d[i];
}
for (rint i = 1; i <= n; i++)
{
double sum = ((double)log2(d[i]) * d[i] + s[n / i]) * phi[i];
ans += sum;
}
cout << fixed << setprecision(10) << ans << endl;
return 0;
}
高洁
前置知识 The Sum of the k-th Powers
求 \((\sum_{i=1}^ni^k) \bmod (10^9+7)\)
答案是一个 \(k+1\) 次多项式，考虑找 \(k+2\) 个值带进去拉格朗日插值
\(n+1\) 组点值\((x_i,y_i)\)，有 \(n\) 次多项式 \(f\) 的拉格朗日插值：
\[f(x) = \sum_{i = 0}^n y_i\prod_{j\not =i} \frac{x-x_j}{x_i-x_j}
\]
复杂度为 \(O(n^2)\).
在此题中把 \(1\) 到 \(k+2\) 带入，有对于每个 \(i\)，分母是\(1\) 乘到 \(i-1\) 再乘上 \(-1\) 乘到 \(i-k-2\)，这可以预处理阶乘 \(O(1)\) 处理。分子可以预处理前后缀积来 \(O(1)\) 得到
复杂度为 \(O(n)\)
signed main()
{
cin >> n >> k;
pl[0] = pr[k + 3] = fac[0] = 1;
for (rint i = 1; i <= k + 2; i++) pl[i] = pl[i - 1] * (n - i) % mod;
for (rint i = k + 2; i >= 1; i--) pr[i] = pr[i + 1] * (n - i) % mod;
for (rint i = 1; i <= k + 2; i++) fac[i] = fac[i - 1] * i % mod;
for (rint i = 1; i <= k + 2; i++)
{
y = (y + qpow(i, k)) % mod;
int a = pl[i - 1] * pr[i + 1] % mod;
int b = fac[i - 1] * ((k - i) & 1 ? -1 : 1) * fac[k + 2 - i] % mod;
ans = (ans + y * a % mod * qpow(b, mod - 2) % mod) % mod;
}
cout << (ans + mod) % mod << endl;
return 0;
}
在看我们的原题
\(v(i)\) 只和 \(\gcd(i,d)\) 有关。枚举 \(d\) 的因数，考虑它的倍数对答案的贡献。
结果为 \(\begin{aligned}\sum_{i\mid d}\sum_{j=1}^{n/i} (ij)^{v(i)}[\gcd(j,d/i)==1]\end{aligned}\)。
整理得：
\(\begin{aligned}\sum_{i\mid d}i^{v(i)+1}\sum_{j=1}^{n/i}j^{v(i)+1}[\gcd(j,d/i)==1] =\sum_{i \mid d}i^{v(i)+1}\sum_{j=1}^{n/i}j^{v(i)+1}\sum_{p \mid i,j}\mu(p) \\=\sum_{i \mid d}i^{v(i)+1}\sum_{p \mid \frac{d}{i}}\mu(p)\sum_{j=1}^{n/i/p} j^{v(i)+1}\end{aligned}\)
\(v(i)\) 为 \(d\) 的所有质因子在 \(d\) 中出现次数与在 \(i\) 中出现次数的商的最大值，可以在枚举 \(i\) 时计算。\(\sum_{i=1}^n i^k\) 前置只是里有的。所以这个题就解决了。
int getpow(int n, int k)
{
n = n % mod;
pl[0] = pr[k + 3] = fac[0] = 1;
int ans = 0, y = 0;
for (rint i = 1; i <= k + 2; i++) pl[i] = pl[i - 1] * (n - i) % mod;
for (rint i = k + 2; i >= 1; i--) pr[i] = pr[i + 1] * (n - i) % mod;
for (rint i = 1; i <= k + 2; i++) fac[i] = fac[i - 1] * i % mod;
for (rint i = 1; i <= k + 2; i++)
{
y = (y + qpow(i, k)) % mod;
int a = pl[i - 1] * pr[i + 1] % mod;
int b = fac[i - 1] * ((k - i) & 1 ? -1 : 1) * fac[k + 2 - i] % mod;
ans = (ans + y * a % mod * qpow(b, mod - 2) % mod) % mod;
}
return (ans + mod) % mod;
}
void solve(int x)
{
memset(c, 0, sizeof c);
for (rint i = 2; i * i <= x; i++)
{
if (!(x % i))
{
p[++cnt] = i;
while (!(x % i))
{
x /= i;
c[cnt]++;
}
}
}
if (x > 1) p[++cnt] = x, c[cnt] = 1;
}
void dfs_init(int x, int step)
{
if (x > cnt)
{
int p2 = 1, flag = 1;
for (rint i = 1; i <= cnt; i++)
{
if (t[i] > 1) flag = 0;
else if (t[i] == 1) flag = -flag;
for (rint j = 1; j <= t[i]; j++) p2 *= p[i];
}
if (!flag) return ;
res2 = (res2 + flag * qpow(p2, step) * getpow(n / p1 / p2, step) % mod) % mod;
return ;
}
for (rint i = 0; i <= c[x] - k[x]; i++)
{
t[x] = i;
dfs_init(x + 1, step);
}
}
void solve(int d, int p1, int step)
{
if (d == 1) step = 1;
step++;
int res = qpow(p1, step);
res2 = 0;
dfs_init(1, step);
ans = (ans + res * res2 % mod) % mod;
}
void dfs(int x)
{
if (x > cnt)
{
p1 = 1;
int maxx = 0;
bool flag = 1;
for (rint i = 1; i <= cnt; i++)
{
flag &= (k[i] > 0);
for (rint j = 1; j <= k[i]; j++) p1 *= p[i];
if (k[i]) maxx = max(maxx, (c[i] + k[i] - 1) / k[i]);
}
solve(d, p1, flag ? maxx : 0);
return ;
}
for (rint i = 0; i <= c[x]; i++)
{
k[x] = i;
dfs(x + 1);
}
}
signed main()
{
fac[0] = 1;
for (rint i = 1; i <= 30; i++) fac[i] = fac[i - 1] * i % mod;
int T;
cin >> T;
while (T--)
{
cin >> n >> d;
cnt = ans = 0;
solve(d);
dfs(1);
cout << (ans % mod + mod) % mod << endl;
}
return 0;
}
理性
把式子展开有一个关于 \(a\) 的二次函数：
\[\sum_{i = 1}^n d_i^2 a^2 + 2(b - v_i)d_i \times a + (b - v_i)^2\\
\]
这个函数的二次项始终为正数，当 \(a\) 取对称轴时函数取到最小值：
\[a = \frac{\sum_{i = 1}^n (v_i - b)d_i}{\sum_{i = 1}^n d_i^2}\\
\left(\sum_{i = 1}^n d_i^2\right)\left(\frac{\sum_{i = 1}^n (v_i - b)d_i}{\sum_{i = 1}^n d_i^2}\right)^2 + 2\left(\sum_{i = 1}^n (b - v_i)d_i\right)\left(\frac{\sum_{i = 1}^n (v_i - b)d_i}{\sum_{i = 1}^n d_i^2}\right) + \left(\sum_{i = 1}^n (b - v_i)^2\right)\\
\frac{(\sum_{i = 1}^n v_id_i)^2 - 2(\sum_{i = 1}^n v_id_i)(b\sum_{i = 1}^n d_i) + b^2(\sum_{i = 1}^n d_i)^2}{\sum_{i = 1}^n d_i^2} + 2\left(b \sum_{i = 1}^n d_i - \sum_{i = 1}^n v_id_i\right)\left(\frac{\sum_{i = 1}^n v_id_i - b \sum_{i = 1}^n d_i}{\sum_{i = 1}^n d_i^2}\right) + \left(\sum_{i = 1}^n b^2 - 2bv_i + v_i^2\right)\\
svd = \sum_{i = 1}^n v_id_i, sd = \sum_{i = 1}^n d_i, sd2 = \sum_{i = 1}^n d_i^2, sv = \sum_{i = 1}^n v_i, sv2 = \sum_{i = 1}^n v_i^2\\
nb^2 - 2sv \times b + sv2 - \frac{b^2 \times sd^2 - 2svd \times sd \times b + svd^2}{sd2}\\
\left(n - \frac{sd^2}{sd2}\right)b^2 + 2\left(\frac{svd \times sd}{sd2} - sv\right)b + sv2 - \frac{svd^2}{sd2}\\
\]
代入后得到了一个只与 \(b\) 有关的二次函数， \(n - \frac{sd^2}{sd2}\) 一定 \(\ge 0\)
所以，再让 \(b\) 取对称轴时函数取到最小值
\[b = \frac{sv - \frac{svd \times sd}{sd2}}{n - \frac{sd^2}{sd2}}\\
sv2 - \frac{svd^2}{sd2} - \frac{(sv - \frac{svd \times sd}{sd2})^2}{n - \frac{sd^2}{sd2}}\\
sv2 - \frac{svd^2}{sd2} - \frac{sv^2 - 2sv\frac{svd \times sd}{sd2} + \left(\frac{svd \times sd}{sd2}\right)^2}{n - \frac{sd^2}{sd2}}\\
sv2 - \frac{svd^2}{sd2} - \left(\frac{sv^2}{n - \frac{sd^2}{sd2}} - 2\frac{\frac{sd}{sd2}}{n - \frac{sd^2}{sd2}}sv \times svd + \frac{\left(\frac{sd}{sd2}\right)^2}{n - \frac{sd^2}{sd2}}svd^2\right)\\
sv2 - \frac{svd^2}{sd2} - \frac{\left(\frac{sd}{sd2}\right)^2}{n - \frac{sd^2}{sd2}}svd^2 - \frac{sv^2}{n - \frac{sd^2}{sd2}} + 2\frac{\frac{sd}{sd2}}{n - \frac{sd^2}{sd2}}sv \times svd\\
\]
得到了对于一个固定的 \(v\)， \(sd, sd2\) 都已知，所以只要求出 \(sv2, svd^2, sv^2, sv \times svd\) 的期望即可。
考虑 \(sv2\)。显然有：
\[sv2 = \sum_{i = 1}^n \frac{S_2(l_i, r_i)}{r_i - l_i + 1}
\]
直接计算即可。
设状态 \(f_{i, 0 / 1 / 2 / 3 / 4}\) 分别表示 \(sv, svd, sv^2, sv \times svd, svd^2\) 只考虑前 \(i\) 项的期望，转移方程：
\[f_{i, 0} = f_{i - 1, 0} + \frac{l_i + r_i}2\\
f_{i, 1} = f_{i - 1, 1} + d_i\frac{l_i + r_i}2\\
f_{i, 2} = f_{i - 1, 2} + (l_i + r_i)f_{i - 1, 0} + \frac{S_2(l_i, r_i)}{r_i - l_i + 1}\\
f_{i, 3} = f_{i - 1, 3} + \frac{l_i + r_i}2f_{i - 1, 1} + d_i\frac{l_i + r_i}2f_{i - 1, 0} + d_i\frac{S_2(l_i, r_i)}{r_i - l_i + 1}\\
f_{i, 4} = f_{i - 1, 4} + d_i(l_i + r_i)f_{i - 1, 1} + d_i^2\frac{S_2(l_i, r_i)}{r_i - l_i + 1}\\
\]
void add(int &x, int y) {x = (x + y) % mod;}
void mul(int &x, int y) {x = x * y % mod;}
int qpow(int a, int b)
{
int res = 1;
while (b)
{
if (b & 1) res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
int inv(int x){return qpow(x, mod - 2);}
int calc(int x){return x * (x + 1) % mod * (x << 1 | 1) % mod * iv6 % mod;}
int calc(int l, int r) {return (calc(r) - calc(l - 1) + mod) % mod;}
signed main()
{
cin >> n;
for (rint i = 1; i <= n; i++)
{
cin >> d[i] >> l[i] >> r[i];
add(sd, d[i]);
add(sd2, d[i] * d[i] % mod);
}
for (rint i = 1; i <= n; i++)
{
int iv = inv(r[i] - l[i] + 1);
add(ans, calc(l[i], r[i]) * iv % mod);
add(f[i][0] = f[i - 1][0], (l[i] + r[i]) * iv2 % mod);
add(f[i][1] = f[i - 1][1], (l[i] + r[i]) * iv2 % mod * d[i] % mod);
add(f[i][2] = f[i - 1][2], calc(l[i], r[i]) * iv % mod);
add(f[i][2], (l[i] + r[i]) * f[i - 1][0] % mod);
add(f[i][3] = f[i - 1][3], calc(l[i], r[i]) * iv % mod * d[i] % mod);
add(f[i][3], (l[i] + r[i]) * f[i - 1][1] % mod * iv2 % mod);
add(f[i][3], (l[i] + r[i]) * f[i - 1][0] % mod * iv2 % mod * d[i] % mod);
add(f[i][4] = f[i - 1][4], calc(l[i], r[i]) * iv % mod * d[i] % mod * d[i] % mod);
add(f[i][4], (l[i] + r[i]) * f[i - 1][1] % mod * d[i] % mod);
}
int q = inv((n - sd * sd % mod * inv(sd2) % mod + mod) % mod);
add(ans, mod - f[n][4] * inv(sd2) % mod);
add(ans, mod - f[n][4] * qpow(sd * inv(sd2) % mod, 2) % mod * q % mod);
add(ans, mod - f[n][2] * q % mod);
add(ans, 2 * f[n][3] * sd % mod * inv(sd2) % mod * q % mod);
cout << ans << endl;
return 0;
}
