数论分块


分块

已知某函数$f(x)$对于$x\in[l,r]$,有$f(x)$关于$x$单调,且$f(x)$值域远小于$x$的定义域。

现在要你求$\sum_{x=1}^n g(x,f(x))$

那么我们就可以根据$f(x)$对$g$进行分块,在这一块中,始终有常数$y=f(x)$,然后对$h(x)=g(x,y)$统计$x$的前缀和。

最终我们就能很快的计算答案。

细节

对于分块,很多时候我们无法直接计算块的范围,需要二分,比如这题 https://nanti.jisuanke.com/t/42386

下面展示详细的二分分块代码:

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int f(ll x) {
__int128 prod = x;
int res = 0;
while (prod * x <= _n) {
prod *= x;
res++;
}
return res;
}

// 找到最大的x, 使得f(x)==f(l)
ll calcEnd(ll l, ll r) {
while (l < r) {
ll mid = (l + r + 1) / 2; // (l,r]
if (f(l) == f(mid)) {
l = mid;
} else {
r = mid - 1;
}
}
return l;
}

// O(lgn * lgn * lgn)
int solve(ll n) {
int ans = 0;
for (ll l = 2, r; l <= n; l = r + 1) {
r = calcEnd(l, n);
int I = f(l);
// printf("cal: %lld %lld %d \n", l, r, I);
int add = (sum(n % mod, I, r) - sum(n % mod, I, l - 1) + mod) % mod;
ans = (ans + add) % mod;
}
return ans;
}

过题代码

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#include <bits/stdc++.h>


using namespace std;
typedef long long ll;

const int maxn = 100;
const int mod = 998244353;

int qpow(int a, int b) {
int ret = 1;
while (b) {
if (b & 1) ret = 1ll * ret * a % mod;
a = 1ll * a * a % mod;
b >>= 1;
}
return ret;
}

// 拉格朗日插值法
int facinv[maxn] = {1, 1};

void facinv_ini() {
for (int i = 0, fac = 1; i < maxn; ++i, fac = 1ll * fac * i % mod) {
facinv[i] = qpow(fac, mod - 2);
}
}

int lagrange(int *y, int n, int x) {// O(n) n次多项式有n+1项 y[0]...y[n] -> y[x]
static int prepre[maxn], suf[maxn], *pre = prepre + 1;
pre[-1] = suf[n + 1] = 1;
for (int i = 0; i <= n; ++i) pre[i] = 1ll * pre[i - 1] * (x - i + mod) % mod;
for (int i = n; i >= 0; i--) suf[i] = 1ll * suf[i + 1] * (i - x + mod) % mod;
int b = 0;
for (int i = 0; i <= n; ++i) {
int up = 1ll * pre[i - 1] * suf[i + 1] % mod;
int down = 1ll * facinv[i] * facinv[n - i] % mod;
b = (b + 1ll * y[i] * up % mod * down) % mod;
}
return b;
}

// O(I)
int sum(int n, int I, ll mx) {
static int f[100];
f[0] = 0;
f[1] = (1ll * (n + 1) * (I + 1) % mod - (I + 1) + mod) % mod;
for (int x = 2; x <= I + 3; x++) {
int left = 1ll * (n + 1) * (I + 1) % mod * x % mod;
int fenzi = (qpow(x, 2) - qpow(x, I + 3) + mod) % mod;
int fenmu = (1 - x + mod) % mod;
f[x] = (left - 1ll * fenzi * qpow(fenmu, mod - 2) % mod + mod) % mod;
}
for (int x = 1; x <= I + 3; x++) {
f[x] = (f[x - 1] + f[x]) % mod;
}
return lagrange(f, I + 3, mx % mod);
}


ll _n;

int f(ll x) {
__int128 prod = x;
int res = 0;
while (prod * x <= _n) {
prod *= x;
res++;
}
return res;
}

// 找到最大的x, 使得f(x)==f(l)
ll calcEnd(ll l, ll r) {
while (l < r) {
ll mid = (l + r + 1) / 2; // (l,r]
if (f(l) == f(mid)) {
l = mid;
} else {
r = mid - 1;
}
}
return l;
}

// O(lgn * lgn * lgn)
int solve(ll n) {
int ans = 0;
for (ll l = 2, r; l <= n; l = r + 1) {
r = calcEnd(l, n);
int I = f(l);
// printf("cal: %lld %lld %d \n", l, r, I);
int add = (sum(n % mod, I, r) - sum(n % mod, I, l - 1) + mod) % mod;
ans = (ans + add) % mod;
}
return ans;
}


int main() {
facinv_ini();

ll n;
while (cin >> n) {
_n = n;
for (int i = 0; i < 1e3; i++) {
solve(n);
}
printf("%d\n", solve(n));
}
}

文章作者: fightinggg
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