數學基礎 5
阿新 • • 發佈：2021-06-27
目錄
數學基礎 5
前言
因為 \(3\) 寫不出來了 \(4\) 現在還空著 但是計劃被打亂 所以來寫 \(5\) (大霧
目前還沒寫完 後面先咕著 學了之後再寫
拉格朗日插值
拉格朗日插值法
眾所周知 給定 \(n + 1\) 個點 可以確定一個 \(n\) 次的多項式
\[f(x) = \sum_{i = 0}^{n}y_i\prod_{i \ne j}\frac{x-x_j}{x_i - x_j} \]求其中某個點對應函式值時 將其代入即可
證明? 沒有證明(我並不會
程式碼
/* Time: 6.27 Worker: Blank_space Source: P4781 【模板】拉格朗日插值 */ /*--------------------------------------------*/ #include<cstdio> #define int long long /*--------------------------------------標頭檔案*/ const int mod = 998244353; int n, k, X[2010], Y[2010], ans; /*------------------------------------變數定義*/ inline int read() { int x = 0, f = 1; char ch = getchar(); while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();} while(ch >= '0' && ch <= '9') {x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar();} return x * f; } /*----------------------------------------快讀*/ int power(int _a, int _b, int res = 1) {for(; _b; _a = _a * _a % mod, _b >>= 1) if(_b & 1) res = res * _a % mod; return res;} /*----------------------------------------函式*/ signed main() { n = read(); k = read(); for(int i = 1; i <= n; i++) X[i] = read(), Y[i] = read(); for(int i = 1; i <= n; i++) { int tmp1 = Y[i], tmp2 = 1; for(int j = 1; j <= n; j++) if(i != j) tmp1 = (tmp1 * (k - X[j]) % mod + mod) % mod, tmp2 = (tmp2 * (X[i] - X[j]) % mod + mod) % mod; ans = (ans + tmp1 * power(tmp2, mod - 2) % mod) % mod; } printf("%lld", ans); return 0; }
在 \(x\) 值連續的時候
將 \(x_i\) 換成 \(i\) 有:
\[f(k) = \sum_{i = 0}^ny_i\prod_{i \ne j}\frac{k - j}{i - j} \]考慮優化後面那一坨東西
分子維護關於 \(k\) 的字首積和字尾積 分母直接用階乘來表示 式子為:
\[f(k) = \sum_{i = 0}^n\left(-1\right)^{n - i}y_i\frac{pre_{i - 1}\times suf_{i + 1}}{fac_i \times fac_{n - i}} \]重心拉格朗日插值法
\[\begin{array}\\ f(x) & = & \sum_{i = 1}^ny_i\prod_{i \ne j}\frac{x - x_j}{x_i - x_j} \\ & = & \sum_{i = 1}^ny_i\frac{\prod_{j \ne i}\left(x - x_j\right)}{\prod_{i \ne j}\left(x_i - x_j\right)}\\ & = & \sum_{i = 1}^ny_i\frac{\prod_{j = 1}^n\left(x - x_j\right)}{\left(x - x_i\right)\prod_{j \ne i}\left(x_i - x_j\right) }\\ & = & \prod_{i = 1}^n\left(x - x_i\right)\sum_{i = 1}^n\frac{y_i}{\left(x - x_i\right)\prod_{i \ne j}\left(x_i - x_i\right)} \end{array} \]令 \(g = \prod_{i = 1}^n\left(x - x_i\right), w(i) = \prod_{j \ne i}\left(x_i - x_j\right)\)
對於每一個新增加的插值點 \(O(n)\) 的更新所有的 \(w(i)\)
程式碼
/* Time: 6.27 Worker: Blank_space Source: */ /*--------------------------------------------*/ #include<cstdio> #define int long long /*--------------------------------------標頭檔案*/ const int mod = 1e9 + 7; int n, X[3010], Y[3010], W[3010], cnt; /*------------------------------------變數定義*/ inline int read() { int x = 0, f = 1; char ch = getchar(); while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();} while(ch >= '0' && ch <= '9') {x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar();} return x * f; } /*----------------------------------------快讀*/ int power(int _a, int _b) {int res = 1; for(; _b; _a = _a * _a % mod, _b >>= 1) if(_b & 1) res = res * _a % mod; return res;} void insert(int x, int y) { X[++cnt] = x; Y[cnt] = y; W[cnt] = 1; for(int i = 1; i < cnt; i++) W[i] = W[i] * (X[i] - x) % mod, W[cnt] = W[cnt] * (x - X[i]) % mod; } void work(int x) { int g = 1, ans = 0; for(int i = 1; i <= cnt; i++) if(X[i] == x) {printf("%lld\n", Y[i]); return ;} else g = g * (x - X[i]) % mod; for(int i = 1; i <= cnt; i++) ans = (ans + Y[i] * power((x - X[i]) * W[i], mod - 2) % mod + mod) % mod; printf("%lld\n", ans * g % mod); } /*----------------------------------------函式*/ signed main() { n = read(); for(int i = 1; i <= n; i++) { int opt = read(), x = read(); if(opt == 1) insert(x, read()); else work(x); } return 0; }
自然數冪和
定義
前 \(n\) 個自然數 \(k\) 次冪的和為
\[S_k(n) = \sum_{i = 1}^ni^k \]性質
\(S_k(n)\) 為關於 \(n\) 的 \(k + 1\) 次的多項式
拉格朗日插值法
需要 \(k + 2\) 個點來計算 可以直接取點 \(O(k^2)\) 的計算
題目對於選擇的點沒有要求 當然可以選取連續的一段 按照上面那個直接帶進去:
\[S_k(n) = \sum_{i = 1}^{k + 2}(-1)^{k + 2 - i}S_k(i)\frac{pre_{i - 1} \times suf_{i + 1}}{fac_{i} \times fac_{k + 2 - i}} \]複雜度: \(O(k\log k)\)
程式碼:
/*
Time: 6.27
Worker: Blank_space
Source: CF622F The Sum of the k-th Powers
*/
/*--------------------------------------------*/
#include<cstdio>
#define int long long
/*--------------------------------------標頭檔案*/
const int C = 1e6 + 7;
const int mod = 1e9 + 7;
/*------------------------------------常量定義*/
int n, k, ans, tmp, pre[C], suf[C], fac[C];
/*------------------------------------變數定義*/
inline int read() {
int x = 0, f = 1; char ch = getchar();
while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
while(ch >= '0' && ch <= '9') {x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar();}
return x * f;
}
/*----------------------------------------快讀*/
int power(int _a, int _b, int res = 1) {for(; _b; _a = _a * _a % mod, _b >>= 1) if(_b & 1) res = res * _a % mod; return res;}
/*----------------------------------------函式*/
signed main() {
n = read(); k = read(); pre[0] = suf[k + 3] = fac[0] = 1;
for(int i = 1; i <= k + 2; i++) pre[i] = pre[i - 1] * (n - i) % mod;
for(int i = k + 2; i >= 1; i--) suf[i] = suf[i + 1] * (n - i) % mod;
for(int i = 1; i <= k + 2; i++) fac[i] = fac[i - 1] * i % mod;
for(int i = 1; i <= k + 2; i++)
{
tmp = (tmp + power(i, k) % mod) % mod;
int x = pre[i - 1] * suf[i + 1] % mod;
int y = ((k - i & 1) ? -1 : 1) * fac[i - 1] * fac[k + 2 - i] % mod;
ans = (ans + tmp * x % mod * power(y, mod - 2) % mod + mod) % mod;
}
printf("%lld", ans);
return 0;
}
其他方法
然而我並不會其他方法(巨霧
題目
首先 由題:
\[ans = \sum_{i = 0}^m\left(\sum_{j = 1}^{n - a_i}j^{m + 1} - \sum_{j = i + 1}^m\left(a_j - a_i \right)^{m + 1} \right) \]然而這個式子我並沒有推出來
然後? 然後就沒有然後了 直接算就好了
程式碼
/*
Time: 6.27
Worker: Blank_space
Source: P4593 [TJOI2018]教科書般的褻瀆
*/
/*--------------------------------------------*/
#include<cstdio>
#include<algorithm>
#define int long long
/*--------------------------------------標頭檔案*/
const int mod = 1e9 + 7;
/*------------------------------------常量定義*/
int T, n, m, k, a[60], ans, tmp, pre[60], suf[60], fac[60];
/*------------------------------------變數定義*/
inline int read() {
int x = 0, f = 1; char ch = getchar();
while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
while(ch >= '0' && ch <= '9') {x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar();}
return x * f;
}
/*----------------------------------------快讀*/
int power(int _a, int _b, int res = 1) {for(; _b; _a = _a * _a % mod, _b >>= 1) if(_b & 1) res = res * _a % mod; return res;}
int Sum(int t) {
pre[0] = suf[k + 3] = fac[0] = 1; tmp = 0; int res = 0;
for(int i = 1; i <= k + 2; i++) pre[i] = pre[i - 1] * (t - i) % mod;
for(int i = k + 2; i >= 1; i--) suf[i] = suf[i + 1] * (t - i) % mod;
for(int i = 1; i <= k + 2; i++) fac[i] = fac[i - 1] * i % mod;
for(int i = 1; i <= k + 2; i++)
{
tmp = (tmp + power(i, k) % mod) % mod;
int x = pre[i - 1] * suf[i + 1] % mod;
int y = ((k - i & 1) ? -1 : 1) * fac[i - 1] * fac[k + 2 - i] % mod;
res = (res + tmp * x % mod * power(y, mod - 2) % mod + mod) % mod;
}
return res;
}
void work() {
n = read(); m = read(); ans = 0; k = m + 1;
for(int i = 1; i <= m; i++) a[i] = read();
std::sort(a + 1, a + 1 + m);
for(int i = 0; i <= m; i++)
{
int sum = Sum(n - a[i]) % mod;
for(int j = i + 1; j <= m; j++) sum = ((sum - power(a[j] - a[i], k) % mod) % mod + mod) % mod;
ans = (ans + sum) % mod;
}
printf("%lld\n", (ans + mod) % mod);
}
/*----------------------------------------函式*/
signed main() {
T = read(); while(T--) work();
return 0;
}
待填