Link BD consists of a single bar 30 mm wide and 12 mm thick. Knowing that each pin has a10 mm diameter, determine the maximum value of the average normal stress in the link BD if θ = 15 degrees.

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 Question

Please help me code this in C++

Sample Input:

1

4

1 2 3 4

Sample Output:

16

Solution

Step 1

Here is the detailed and simplified C++ code for the above listed problem statement:

Step 2

C++ Code:

 

#include <bits/stdc++.h>
#define fastIo ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL)
#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
#define fi first
#define se second
#define sz size
#define pb push_back
#define mp make_pair
using namespace std;

const double EPS = 1e-9;
const double PI = 3.141592653589793238462;

const int dr[] = {1, 1, 0, -1, -1, -1, 0, 1};
const int dc[] = {0, 1, 1, 1, 0, -1, -1, -1};

const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};

typedef long long ll;

ll a[65], minR[65][65], sum[65][65];
ll dp[65][65];

ll getMin(int i, int j) {
  if (i > j) return -1000000000LL * 60LL - 1LL;
  return minR[i][j];
}

ll solve(int lef, int rig) {
  if (lef + 1 == rig) {
    return minR[lef][rig] * 2;
  }
  if (dp[lef][rig] != -1000000000LL * 60LL - 1LL) {
    return dp[lef][rig];
  }
  ll ret = minR[lef][rig] * (rig - lef + 1);
  for (int l = lef + 1; l < rig; l++) {
    ret = max(ret, solve(l, rig) + min(minR[lef][l - 1], sum[l][rig]) * ((l - 1) - lef + 1 + 1));
    ret = max(ret, solve(lef, l) + min(minR[l + 1][rig], sum[lef][l]) * (rig - (l + 1) + 1 + 1));
    for (int r = l + 1; r < rig; r++) {
      ll cur = solve(lef, l) + solve(r, rig);
      if (r == l + 1) {
        ll add = min(sum[lef][l], sum[r][rig]);
        ret = max(ret, cur + 2 * add);
      } else {
        ll add = min(sum[lef][l], min(sum[r][rig], getMin(l + 1, r - 1)));
        ret = max(ret, cur + (r - 1 - (l + 1) + 1 + 2) * add);
      }
    }
  }
  return dp[lef][rig] = ret;
}

void solve() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; i++) {
    minR[i][i] = sum[i][i] = a[i];
    for (int j = i + 1; j <= n; j++) {
      minR[i][j] = min(minR[i][j - 1], a[j]);
      sum[i][j] = sum[i][j - 1] + a[j];
    }
  }
  for (int i = 1; i < 65; i++) {
      for (int j = 1; j < 65; j++) {
          dp[i][j] = -1000000000LL * 60LL - 1LL;
      }
  }
  cout << solve(1, n) << '\n';
}
int main() {
  fastIo;
  int t; cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

 

Output: