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类欧几里得算法详细推导过程（附带模板）

發布時間：2023/12/4 编程问答 31 豆豆

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類歐幾里得算法推導

初識

給出三種形式：

$\sum_{i = 0} ^{n} \lfloor\frac{ai + b}{c}\rfloor$
$\sum_{i = 0} ^{n}i \lfloor \frac{ai + b}{c}\rfloor$
$\sum_{i = 1} ^{n} \lfloor \frac{ai + b}{c}\rfloor ^ 2$

接下來我們進入正題，講解這三個式子的求解

f(a, b, c, n)

$\sum_{i = 1} ^{n} \lfloor \frac{ai + b}{c} \rfloor\\ if\ a \geq c\ 或者 b \geq c\\ f(a, b, c, n) = f(a \% c, b \% c, c, n) + \frac{n(n + 1)}{2} \lfloor\frac{a}{c}\rfloor + (n + 1) \lfloor \frac{b}{c}\rfloor\\ if\ a < c 并且 b < c\\ 我們假定m = \lfloor\frac{an + b}{c}\rfloor\\ 有f(a, b, c, n) = \sum_{i = 0} ^{n} \sum_{j = 1} ^{m}(\lfloor\frac{ai + b}{c}\rfloor \geq j)\\ = \sum_{i = 0} ^{n} \sum_{j = 0} ^{m - 1} (\lfloor\frac{ai + b}{c}\rfloor \geq j + 1)\\ = \sum_{i = 0} ^{n} \sum_{j = 0} ^{m - 1} (i > \lfloor\frac{jc + c - b - 1}{a})\rfloor\\ = \sum_{j = 0} ^{m - 1} n - \lfloor\frac{jc + c - b - 1}{a}\rfloor\\ = nm - f(c, c - b - 1, a, m - 1)\\ 也就是a = c, c = a \% c\\ 最后的遞歸邊界就是a = 0\\$
最后我們就可以通過遞歸求得其函數值。

g(a, b, c, n)

$\sum_{i = 0} ^{n} i\lfloor \frac{ai + b}{c}\rfloor\\ if\ a \geq c 或者 b \geq c\\ g(a, b, c, n) = g(a \%c, b\%c, c, n) + \frac{n(n + 1)(2n + 1)}{6} \lfloor\frac{a}{c}\rfloor + \frac{n(n + 1)}{2} \lfloor\frac{b}{c}\rfloor\\ if\ a < c 并且 b < c\\ 同上面式子一樣m = \lfloor\frac{an + b}{c}\rfloor\\ g(a, b, c, n) = \sum_{i = 0} ^{n}i \sum_{j = 1} ^{m} (\lfloor\frac{ai + b}{c}\rfloor \geq j)\\ = \sum_{i = 0} ^{n} i\sum_{j = 0} ^{m - 1} (i > \lfloor\frac{jc + c - b - 1}{a})\rfloor\\ 顯然有i的下界為\lfloor\frac{jc + c - b - 1}{a}\rfloor + 1，上界是n，所以構成了一個等差數列\\ 最后得到g(a, b, c, n) = \sum_{j = 0} ^{m - 1} \lfloor\frac{(\lfloor\frac{jc + c - b - 1}{a}\rfloor + 1 + n)(n - \lfloor\frac{jc + c - b - 1}{a}\rfloor)}{2}\rfloor\\ = \lfloor\frac{mn(n + 1) - f(c, c - b - 1, a, m - 1) - h(c, c - b - 1, a, m - 1)}{2}\rfloor\\ 所以我們先推導h(a, b, c, n)\\$

h(a, b, c, n)

$\sum_{i = 0} ^{n} \lfloor\frac{ai + b}{c}\rfloor ^ 2\\ if\ a \geq c\ 或者 b \geq c\\ h(a, b, c, n) = \sum_{i = 0} ^{n}\left( \frac{i(a \% c) + b \% c}{c} + \lfloor\frac{a}{c}\rfloor i + \lfloor \frac{b}{c}\rfloor \right)^ 2\\ = \sum_{i = 0} ^{n} \left( \frac{i(a \% c) + b \% c}{c}\right)^ 2 + \sum_{i = 0} ^{n} \lfloor\frac{a}{c}\rfloor ^ 2 i ^ 2 + \sum_{i = 0} ^{n} \lfloor \frac{b}{c}\rfloor ^ 2 + 2\lfloor\frac{a}{c}\rfloor \sum_{i = 0} ^{n} i \frac{i(a \% c) + b \% c}{c} + 2 \lfloor\frac{b}{c}\rfloor \sum_{i = 0} ^{n}\frac{i(a \% c) + b \% c}{c} + 2\lfloor\frac{a}{c}\rfloor \lfloor \frac{b}{c}\rfloor \sum_{i = 0} ^{n}i \\ = h(a\%c, b \%c, c, n) + 2 \lfloor \frac{b}{c}\rfloor f(a \%c,b \%c,c,n) +2 \lfloor\frac{a}{c}\rfloor g(a \%c, b\%c, c, n) + \frac{n(n + 1)(2n + 1)}{6}\lfloor\frac{a}{c} \rfloor ^ 2 +n(n + 1) \lfloor\frac{a}{c}\rfloor \lfloor \frac{b}{c}\rfloor + (n + 1) \lfloor\frac{b}{c}\rfloor ^ 2\\ if\ a < c 并且 b < c\\ n ^ 2 = 2 \frac{n(n + 1)}{2} - n = (2\sum_{i = 0} ^{n} i )- n\\ h(a, b, c, n) = \sum_{i = 0} ^{n} (2\sum_{j = 0} ^{\lfloor \frac{ai + b}{c}\rfloor} j) - \lfloor \frac{ai + b}{c}\rfloor\\ m = \lfloor \frac{an + b}{c} \rfloor\\ h(a, b, c, n) = 2 \sum_{j = 0} ^{m - 1}(j + 1) \sum_{i = 0} ^{n} (\lfloor \frac{ai + b}{c}\rfloor \geq j + 1) - f(a, b, c, n)\\ = 2\sum_{j = 0} ^{m} (j + 1) \sum_{i = 0} ^{n} i > \lfloor \frac{cj + c-b-1}{a}\rfloor= \sum_{j = 0} ^{n}(j + 1)(n - \lfloor \frac{cj + c-b-1}{a})\rfloor) - f(a, b, c, n) \\ = 2\left( \sum_{j = 1} ^{m} jn - \sum_{j = 0} ^{m - 1}(j + 1)\lfloor \frac{cj + c-b-1}{a})\rfloor \right) - f(a, b, c, n) \\ 最后得到h(a, b, c, n) = nm(m + 1) - 2f(c, c - b - 1, a, m - 1) - 2g(c, c - b - 1, a, m - 1)- f(a, b, c, n)\\$

總結

$f (a, b, c, n)$ 是可以單獨求的，但是 $g (a, b, c, n), h (a, b, c, n)$ 是得互相得到并且要通過 $f (a, b, c, n)$ 得到，所以我們可以通過一個結構體一次求得三個函數的值，然后一起返回。

P5170 【模板】類歐幾里得算法

/*Author : lifehappy */ #pragma GCC optimize(2) #pragma GCC optimize(3) #include <bits/stdc++.h>#define mp make_pair #define pb push_back #define endl '\n' #define mid (l + r >> 1) #define lson rt << 1, l, mid #define rson rt << 1 | 1, mid + 1, r #define ls rt << 1 #define rs rt << 1 | 1using namespace std;typedef long long ll; typedef unsigned long long ull; typedef pair<int, int> pii;const double pi = acos(-1.0); const double eps = 1e-7; const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x; }const int mod = 998244353, inv2 = 499122177, inv6 = 166374059;struct Node {ll f, g, h; };ll quick_pow(ll a, ll n) {ll ans = 1;while(n) {if(n & 1) ans = ans * a % mod;a = a * a % mod;n >>= 1;}return ans; }Node solve(ll a, ll b, ll c, ll n) {Node ans, temp;if(!a) {ans.f = (n + 1) * (b / c) % mod;ans.g = (n + 1) * n % mod * inv2 % mod * (b / c) % mod;ans.h = (n + 1) * (b / c) % mod * (b / c) % mod;return ans;}if(a >= c || b >= c) {temp = solve(a % c, b % c, c, n);ans.f = (temp.f + n * (n + 1) % mod * inv2 % mod * (a / c) % mod + (n + 1) * (b / c) % mod) % mod;ans.g = (temp.g + n * (n + 1) % mod * (2 * n + 1) % mod * inv6 % mod * (a / c) % mod + n * (n + 1) % mod * inv2 % mod * (b / c) % mod) % mod;ans.h = (temp.h + 2 * (b / c) % mod * temp.f % mod + 2 * (a / c) % mod * temp.g % mod + n * (n + 1) % mod * (2 * n + 1) % mod * inv6 % mod * (a / c) % mod * (a / c) % mod + n * (n + 1) % mod * (a / c) % mod * (b / c) % mod + (n + 1) * (b / c) % mod * (b / c) % mod) % mod;return ans;}ll m = (a * n + b) / c;temp = solve(c, c - b - 1, a, m - 1);ans.f = (n * m % mod - temp.f + mod) % mod;ans.g = ((m * n % mod * (n + 1) % mod - temp.f - temp.h) % mod * inv2 % mod + mod) % mod;ans.h = ((n * m % mod * (m + 1) % mod - 2 * temp.f % mod - 2 * temp.g % mod - ans.f) % mod + mod) % mod;return ans; }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int T = read();while(T--) {ll n = read(), a = read(), b = read(), c = read();Node ans = solve(a, b, c, n);printf("%lld %lld %lld\n", ans.f, ans.h, ans.g);}return 0; }

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