CF1418E. Expected Damage

发表于 | 阅读数

不想写题意

一定要单独考虑。

这种排列组合，一定要考虑概率,首先$\leq b$一定可以$1-\frac{a}{big}$

然后小的一定要单独考虑$big+1$个间隙不能选择$a$种。

一定要单独考虑

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 = 1e6 + 10;
const int mod = 998244353;
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) { return (x + y >= mod) ? (x + y - mod) : (x + y); }
int del(int x, int y) { return (x - y < 0) ? (x - y + mod) : (x - y); }
int qpow(int a, int x, int mo)
{
    int res = 1;
    while (x)
    {
        if (x & 1)
            res = 1ll * a * res % mo;
        a = 1ll * a * a % mo;

        x >>= 1;
    }
    return res;
}
int d[N];
int sum[N];
int main()
{
    int n = read(), m = read();
    for (int i = 1; i <= n; i++)
        d[i] = read();
    sort(1 + d, 1 + d + n);
    for (int i = 1; i <= n; i++)
        sum[i] = inc(sum[i - 1], d[i] % mod);
    for (int i = 1; i <= m; i++)
    {
        int a = read(), b = read();
        int k = lower_bound(1 + d, 1 + d + n, b) - d;
        k--;
        int big = n - k;
        if (big < a)
        {
            puts("0");
        }
        else
        {
            int ans = 0;

            ans = inc(1ll * del(sum[n], sum[k]) * (big - a) % mod * qpow(big, mod - 2, mod) % mod, ans);
            ans = inc(1ll * sum[k] * (big + 1 - a) % mod * qpow(big + 1, mod - 2, mod) % mod, ans);
            cout << ans << endl;
        }
    }
}