2019 ICPC 邀请赛(南昌) B-Polynomial

发表于 | 阅读数

已知$f_0..f_n$,求$\sum_l^r f(i)$

傻逼卡常题，模版题。预处理下阶乘和连乘的东西就可少个$log$。

代码 
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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mk make_pair
const int N = 2e6 + 10;
const int mod = 9999991;

int read()
{
    int x = 0, f = 1;
    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 - '0', c = getchar();
    return x * f;
}
int inc(int x, int y)
{
    if (y < 0)
        y += mod;
    if (x + y >= mod)
        x -= mod;
    return x + y;
}
int qpow(int a, int x, int mo)
{
    int res = 1;
    while (x)
    {
        if (x & 1)
            res = 1ll * res * a % mo;
        x >>= 1;
        a = 1ll * a * a % mo;
    }
    return res;
}

int fac[N];
int pre[N], suf[N];
int get(int *y, int x, int len)
{
    if (x < 0)
        return 0;
    if (x <= len)
        return y[x];
    int ans = 0;
    pre[0] = suf[len] = 1;
    for (int i = 1; i <= len; ++i)
        pre[i] = 1ll * pre[i - 1] * (x - i + 1) % mod;
    for (int i = len; i >= 1; --i)
        suf[i - 1] = 1ll * suf[i] * (x - i) % mod;
    for (int i = 0; i <= len; i++)
    {
        int s1 = 1ll * pre[i] * suf[i] % mod;
        int s2 = 1ll * fac[i] * fac[len - i] % mod;
        if ((len - i) & 1)
            s2 = (mod - s2) % mod;
        //cout << s1 << " " << s2 << endl;
        ans = inc(ans, 1ll * y[i] * s1 % mod * s2 % mod);
    }
    return ans;
}
int y[N];
int f[N];
int n, m;
int main()
{
    int T = read();
    fac[0] = 1;
    for (int i = 1; i <= 3000; i++)
        fac[i] = 1ll * fac[i - 1] * i % mod;
    for (int i = 1; i <= 3000; i++)
        fac[i] = qpow(fac[i], mod - 2, mod);
    while (T--)
    {
        n = read(), m = read();

        for (int i = 0; i <= n; i++)
            f[i] = read();
        f[n + 1] = get(f, n + 1, n);
        //cout << f[n + 1] << endl;
        y[0] = f[0];
        for (int i = 1; i <= n + 1; i++)
            y[i] = inc(y[i - 1], f[i]);

        while (m--)
        {
            int L = read(), R = read();
            printf("%d\n", inc(get(y, R, n + 1), -get(y, L - 1, n + 1)));
        }
    }
}