宝塔服务器面板，一键全能部署及管理，送你10850元礼包，点我领取
适用范围：给出n个点$(x_i,y_i)$，过这n个点能够确定一个最高n-1次的多项式$f(x)$，求$f(k)$
做法：如图所示，我们将每一个点$(x_i,y_i)$在x轴上的投影$(x_i,0)$记为$H_i$。对每一个i，我们选择一个点集$lbrace P_ibrace cup lbrace H_j vert 1 le ile n, j
eq ibrace$，求过这n个点的最高n-1次的多项式$g_i(x)$。这样我们就得到了n个$g_i(x)$，它们都在各自对应的$x_i$处的值为$y_i$，并且在其它$x_j(i
eq j)$处值为0。
很容易就能够构造出$g_i(x)$的表达式：
$$g_i(x)=y_i*prod_{j
eq i}frac{x-x_j}{x_i-x_j}$$
显然最后有：
$$f(x)=sum_{i=1}^{n}g_i(x)=sum _{i=1}^{n}y_i*prod_{j
eq i}frac{x-x_j}{x_i-x_j}$$
由于只用求$f(k)$的值，代入得$f(k)=sum _{i=1}^{n}y_i*prod_{j
eq i}frac{k-x_j}{x_i-x_j}$
例题：
#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> using namespace std; typedef long long ll; const ll mod = 998244353; const int N = 2010; int n; ll k, x[N], y[N]; ll power(ll a, ll n) { ll res = 1; while (n) { if (n & 1) res = res * a % mod; a = a * a % mod; n >>= 1; } return res; } int main() { scanf("%d%lld", &n, &k); for (int i = 1; i <= n; i++) scanf("%lld%lld", &x[i], &y[i]); ll res = 0; for (int i = 1; i <= n; i++) { ll ta = y[i] % mod, tb = 1; for (int h = 1; h <= n; h++) { if (h == i) continue; ta = (ta * ((k - x[h]) % mod + mod) % mod) % mod; tb = (tb * ((x[i] - x[h]) % mod + mod) % mod) % mod; } res = (res + ta * power(tb, mod - 2) % mod) % mod; } printf("%lld ", res); return 0; }
[模板]拉格朗日插值
Codeforces – 622F The Sum of the k-th Powers
题意：求$sum_{i=1}^{n}i^k,1leq nleq 10^9,0leq k leq 10^6$
思路：首先有一个结论，$sum_{i=1}^{n}i^k$为k+1阶多项式，所以我们只需要暴力算出$f(n)=sum_{i=1}^{n}i^k$的前k+2项，然后用拉格朗日插值法求第n项即可
下面简单证明一下$sum_{i=1}^{n}i^k$为k+1阶多项式
对于一个数列${a_n}$来说，把数列${a_n}$的元素两两做差得到数列${b_n}$，我们称数列${b_n}$为数列${a_n}$的一阶阶差数列，如果再将数列${b_n}$的元素两两做差得到数列${c_n}$，我们称数列${c_n}$为数列${a_n}$的一阶阶差数列，依次类推定义p阶阶差数列
定理：数列${a_n}$是一个p阶等差数列的充要条件是数列的通项$a_n$为n的一个p次多项式
证明：设$f(x)=sum_{i=0}^{n}u_i x^i$，令${b_n}$为${a_n}$的一阶阶差数列
那么$Delta f(x)=f(x+1)-f(x)=sum_{i=0}^{n}u_i (x+1)^i-sum_{i=0}^{n}u_i x^i$
我们只考虑$x^n$有$Delta f(x)=u_n(x+1)^n-u_n x^n$
将$u_n(x+1)^n$二项式展开，仅考虑$x^n$有$Delta f(x)=u_n x^n-u_n x^n=0$
所以每次差分后多项式的最高次数会减1，p次差分后，变为常数项，此时多项式的最高次数为0，定理成立
显然，如果一个多项式k次差分之后变为0，那么这个多项式的最高次数为k-1
我们将$sum_{i=1}^{n}i^k$差分后得$1^k,2^k,3^kdots n^k$
显然$1^k,2^k,3^kdots n^k$为k阶多项式，所以$sum_{i=1}^{n}i^k$为k+1阶多项式
#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> using namespace std; typedef long long ll; const int N = 2000010; const ll mod = 1000000007; int n, k; ll a[N], f[N], nf[N], now; ll power(ll a, ll n) { ll res = 1; while (n) { if (n & 1) res = res * a % mod; a = a * a % mod; n >>= 1; } return res; } ll solve() { now = f[0] = nf[0] = 1; for (int i = 1; i <= k + 2; i++) { now = now * (n - i) % mod; f[i] = f[i - 1] * i % mod; nf[i] = -nf[i - 1] * i % mod; } ll res = 0; for (int i = 1; i <= k + 2; i++) { ll t = power(f[i - 1] * nf[k + 2 - i] % mod, mod - 2); res = (res + a[i] * now % mod * power(n - i, mod - 2) % mod * t % mod) % mod; res = (res + mod) % mod; } return res; } int main() { // freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); scanf("%d%d", &n, &k); for (int i = 1; i <= min(k + 2, n); i++) a[i] = (a[i - 1] + power(i, k)) % mod; if (n <= k + 2) printf("%lld ", a[n]); else printf("%lld ", solve()); return 0; }
The Sum of the k-th Powers
题意：求$sum_{i=0}^m f(n-a_i)-sum_{i=0}^{m-1} sum_{j=i+1}^{m}(a_j-a_i)^m+1$，其中$f(n)=sum_{i=1}^{n} i^{m+1}$
思路：后半部分$sum_{i=0}^{m-1} sum_{j=i+1}^{m}(a_j-a_i)^m+1$可以直接暴力求解
$f(n)=sum_{i=1}^{n} i^{m+1}$为m+2阶多项式，考虑到m不是很大，所以我们可以插入m+3个值后，暴力对每一个$a_i$求一次f(n-a_i)，最后相加即可
#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> using namespace std; typedef long long ll; const int N = 60; const ll mod = 1000000007; int T; ll n, m, now; ll a[N], c[N], f[N], nf[N]; ll power(ll a, ll n) { ll res = 1; while (n) { if (n & 1) res = res * a % mod; a = a * a % mod; n >>= 1; } return res; } ll solve(ll n) { now = f[0] = nf[0] = 1; for (int i = 1; i <= m + 3; i++) { now = now * (n - i) % mod; f[i] = f[i - 1] * i % mod; nf[i] = -nf[i - 1] * i % mod; } ll res = 0; for (int i = 1; i <= m + 3; i++) { ll t = power(f[i - 1] * nf[m + 3 - i] % mod, mod - 2); res = (res + c[i] * now % mod * power(n - i, mod - 2) % mod * t % mod) % mod; res = (res + mod) % mod; } return res; } ll calc(ll x) { if (x <= m + 3) return c[x]; return solve(x); } int main() { // freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); scanf("%d", &T); while (T--) { scanf("%lld%lld", &n, &m); for (int i = 1; i <= m; i++) scanf("%lld", &a[i]); sort(a + 1, a + m + 1); for (int i = 1; i <= min(m + 3, n); i++) c[i] = (c[i - 1] + power(i, m + 1)) % mod; ll res = 0; for (int i = 0; i <= m; i++) res = (res + calc(n - a[i])) % mod; for (int i = 0; i <= m - 1; i++) { for (int h = i + 1; h <= m; h++) { ll t = power(a[h] - a[i], m + 1); res = ((res - t) % mod + mod) % mod; } } printf("%lld ", res); } return 0; }
[TJOI2018]教科书般的亵渎
题意：求$sum_{i=0}^n sum_{j=1}^{a+i*d} sum_{x=1}^{j}x^k$
思路：$f(n)=sum_{x=1}^{n}x^k$，为k+1阶多项式
$g(n)=sum_{i=1}^{n} sum_{j=1}^{i} j^k$，g(n)差分后的结果为f(n)，所以g(n)为k+2阶多项式
$res(n)=sum_{i=0}^{n}g(a+i*d)$，res(n)进行k+5次差分后为0，所以res(n)为k+4阶多项式
然后插值求解即可
#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> using namespace std; typedef long long ll; const int N = 200; const ll mod = 1234567891; int T, k; ll a, n, d, now, fc[N], nfc[N]; ll f[N], g[N]; ll power(ll a, ll n) { ll res = 1; while (n) { if (n & 1) res = res * a % mod; a = a * a % mod; n >>= 1; } return res; } ll solve(ll n, int m, ll *c) { now = fc[0] = nfc[0] = 1; for (int i = 1; i <= m; i++) { now = now * (n - i) % mod; fc[i] = fc[i - 1] * i % mod; nfc[i] = -nfc[i - 1] * i % mod; } ll res = 0; for (int i = 1; i <= m; i++) { ll t = power(fc[i - 1] * nfc[m - i] % mod, mod - 2); res = (res + c[i] * now % mod * power(n - i, mod - 2) % mod * t % mod) % mod; res = (res + mod) % mod; } return res; } ll calc(ll n, int m, ll *c) { if (n <= m) return c[n]; return solve(n, m, c); } int main() { // freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); scanf("%d", &T); while (T--) { scanf("%d%lld%lld%lld", &k, &a, &n, &d); for (int i = 0; i <= k + 3; i++) f[i] = (f[i - 1] + power(i, k)) % mod; for (int i = 1; i <= k + 3; i++) f[i] = (f[i] + f[i - 1]) % mod; for (int i = 0; i <= k + 5; i++) g[i] = calc((a + i * d) % mod, k + 3, f); for (int i = 1; i <= k + 5; i++) g[i] = (g[i] + g[i - 1]) % mod; printf("%lld ", calc(n, k + 5, g)); } return 0; }
tyvj 1858 XLkxc
支付宝扫一扫
微信扫一扫
