使用c++设计：有一个数组a有n个元素，把数组中的0等概率替换成1~m的某个数，然后对数组排序，计算第k个位置的数，就是ak,计算ak的期望(不能使用rand（）函数，确切的计算出来)，并对结果mod998244353，输入是n,m,k,以及数组的n个值,使用O(n^2logn)以内的算法设计，直接写代码就可以了

#include <stdio.h>
#include <stdlib.h>

#define MOD 998244353

int n, m, k;
int a[100005];

int binom[100005][505]; // combinatorics: binom[n][k] = n choose k

// precomputes binom coefficients
void precompute_binom() {
for (int i = 0; i <= n; i++) {
binom[i][0] = binom[i][i] = 1;
for (int j = 1; j < i; j++) {
binom[i][j] = (binom[i-1][j-1] + binom[i-1][j]) % MOD;
}
}
}

int main() {
scanf("%d %d %d", &n, &m, &k);

precompute_binom();

int zero_count = 0;
for (int i = 0; i < n; i++) {
scanf("%d", &a[i]);
if (a[i] == 0) zero_count++;
}

// compute expected value of each position
long long expected[100005] = {0};
for (int i = 0; i <= zero_count; i++) { // number of zeros replaced
for (int j = 1; j <= m; j++) { // value to replace with
int p = binom[zero_count][i]; // number of ways to choose i zeros to replace
p = 1LL * p * binom[n-zero_count][k-i-1] % MOD; // number of ways to choose k-i-1 other elements
expected[k] = (expected[k] + 1LL*p*j) % MOD; // add to expected value of kth position
}
}

// sort the array
qsort(a, n, sizeof(int), [](const void* a, const void* b) -> int {
return *(int*)a - *(int*)b;
});

// output answer
printf("%lld\n", expected[k]);

return 0;
}