7/16 训练笔记

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插，就硬插，插完就过了（

P4781 【模板】拉格朗日插值

模板题，写拉格朗日插值即可。
代码：

#include <bits/stdc++.h>
#define int long long
#define rep(i, l, r) for (int i = l; i <= r; i++)
using namespace std;
const int mod = 998244353;
int x[2010], y[2010], n, k;
int qpow(int x, int y) {
    int res = 1;
    while (y) {
        if (y & 1) (res *= x) %= mod;
        (x *= x) %= mod;
        y >>= 1;
    }
    return res;
}
void lagrange() {
    int ans = 0;
    rep (i, 1, n) {
        int res1 = 1, res2 = 1;
        rep (j, 1, n) {
            if (i == j) continue;
            (res1 *= ((k - x[j]) % mod + mod) % mod) %= mod;
            (res2 *= ((x[i] - x[j]) % mod + mod) % mod) %= mod;
        }
        (ans += (y[i] * res1 % mod) * qpow(res2, mod - 2) % mod) %= mod;
    }
    cout << (ans + mod) % mod << "\n";
}
signed main() {
    cin >> n >> k;
    rep (i, 1, n) {
        cin >> x[i] >> y[i];
    }
    lagrange();
}

P4593 [TJOI2018] 教科书般的亵渎

注意到答案是一大堆形如 \(\Sigma_{i = 1}^n i^k\) 的项加起来，那么拉格朗日插值。
代码：

#include <bits/stdc++.h>
#define int long long
#define rep(i, l, r) for (int i = l; i <= r; i++)
using namespace std;
const int mod = 1e9 + 7;
int x[2010], y[2010], a[60], t, n, m, k, cnt, ans;
int qpow(int x, int y) {
    int res = 1;
    while (y) {
        if (y & 1) (res *= x) %= mod;
        (x *= x) %= mod;
        y >>= 1;
    }
    return res;
}
int lagrange() {
    int ans = 0;
    rep (i, 1, cnt + 2) {
        int res1 = 1, res2 = 1;
        rep (j, 1, cnt + 2) {
            if (i == j) continue;
            (res1 *= ((k - x[j]) % mod + mod) % mod) %= mod;
            (res2 *= ((x[i] - x[j]) % mod + mod) % mod) %= mod;
        }
        (ans += (y[i] * res1 % mod) * qpow(res2, mod - 2) % mod) %= mod;
    }
    return (ans + mod) % mod;
}
signed main() {
    cin >> t;
    while (t--) {
        cin >> n >> m;
        rep (i, 1, m) {
            cin >> a[i];
        }
        cnt = m + 1;
        rep (i, 1, m + 3) {
            x[i] = i;
            y[i] = (y[i - 1] + qpow(i, cnt)) % mod;
        }
        sort(a + 1, a + m + 1);
        ans = 0;
        rep (i, 1, m + 1) {
            int len = a[i - 1] + 1;
            k = n - len + 1;
            (ans += lagrange()) %= mod;
            rep (j, i, m) ans = ((ans - qpow(a[j] - len + 1, cnt)) % mod + mod) % mod;
        }
        cout << ans << "\n";
    }
}