D. Shuffle（cf）暴力枚举 + 组合数学

原题链接：Problem - 1622D - Codeforces

题目大意：给你一串01串，告诉你长度n和一个数k，这个串中所有1的数量为k的子串，可以把这段子串重新排序。问你最后这个串能有多少种。

AC代码：

#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
typedef pair<int, int> PII;
const double pi = acos(-1.0);
#define rep(i, n) for (int i = 1; i <= (n); ++i)
#define rrep(i, n) for (int i = n; i >= (1); --i)
typedef long long ll;
#define sqar(x) ((x)*(x))

const ll M = 998244353;
const int N = 5010;
char s[N];
int a[N], pre[N];
ll fac[N], invfac[N];

inline ll qpow(ll a, ll n, ll p)// 快速幂
{
    ll ans = 1;
    while (n)
    {
        if (n & 1)
            ans = ans % p * a % p;
        a = a % p * a % p;
        n >>= 1;
    }
    return ans;
}

inline ll inv(ll a, ll p)
{
    return qpow(a, p - 2, p);
}

void init() //注意要在主函数中init()!!!
{
     fac[0] = 1;
     for (int i = 1; i < N; ++i) fac[i] = (fac[i - 1] * i) % M;
     invfac[N - 1] = inv(fac[N - 1], M);
     for (int i = N - 2; i >= 0; --i) invfac[i] = (invfac[i + 1] * (i + 1)) % M;
}

ll C(int n, int m)
{
    if (n < m || m < 0) return 0;
    return fac[n] * invfac[m] % M * invfac[n - m] % M;
}

int main()
{
    init();
    int n, k;
    scanf("%d %d", &n, &k);
    scanf("%s", s + 1);
    rep(i, n) a[i] = s[i] - '0';
    rep(i, n) pre[i] = pre[i - 1] + a[i];
    ll ans = 1;
    if(pre[n] < k){puts("1"); return 0;}
    rep(i, n)
     for(int j = i + 1; j <= n; j++){
        if(pre[j] - pre[i - 1] > k) break;
        int cnt = pre[j] - pre[i - 1];
        cnt -= (a[i] == 0) + (a[j] == 0);
        ans = (ans + C(j - i - 1, cnt)) % M;
     }
    printf("%lld", ans);
    return 0;
}