这题有点意思，首先要发现递增 $p_0$ 总是最优的：设 $c_j = \frac{p_j}{preSum_{j-1}}$，如果递增的是 $p_i$，如果 $i > j$，此时 $c_j$ 不变；如果 $i = j$，此时 $c_j$ 变大；如果 $i < j$，那么 $c_j$ 变小。我们希望对某个元素 $p_j$ 递增的时候，之前的性质能够保持不变，所以每次都递增 $p_0$ 即可。判定整个数列满足题目要求的性质是线性的，所以可以二分答案。

// Date: Sun Dec  3 17:54:52 2023

#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>

#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

using namespace std;

const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double eps = 1e-8;
const int dir[8][2] = {
    {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1},
};

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;

const ull Pr = 131;

#define For(i, a, b) for (int i = int(a); i < int(b); ++i)
#define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i)
#define For1(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i)
#define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i])

#define f1 first
#define f2 second
#define pb push_back
#define has(a, x) (a.find(x) != a.end())
#define nonempty(a) (!a.empty())
#define all(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())

#ifdef _DEBUG
#define debug1(x) cout << #x " = " << x << endl;
#define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl;
#define debug3(x, y, z)                                                        \
  cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl;
#else
#define debug1
#define debug2
#define debug3
#endif

const int N = 110;

int t;

ll n, k, p[N], b[N];

bool check(ll x) {
  For1(i, 2, n) {
    ll l = 100 * p[i], r = k * (x + b[i - 1]);
    if (l > r)
      return false;
  }
  return true;
}

int main(void) {
#ifdef _DEBUG
  freopen("1476b.in", "r", stdin);
#endif
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  cin >> t;

  while (t--) {
    cin >> n >> k;
    memset(b, 0, sizeof b);

    For1(i, 1, n) {
      cin >> p[i];
      b[i] = b[i - 1] + p[i];
    }

    ll l = 0, r = 1e11 + 10, mid;

    while (l < r) {
      mid = (l + r) / 2;
      if (check(mid))
        r = mid;
      else
        l = mid + 1;
    }

    cout << r << '\n';
  }

  return 0;
}