Minimum Replacements to Sort the Array LeetCode solution in C++

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class Solution {
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 public:
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  long long minimumReplacement(vector<int> &nums) {
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    long long res = 0;
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    int n = nums.size();
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    int mxm = nums[n - 1];
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    long long val;
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    for (int i = n - 2; i >= 0; i--) {
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      // minimum no. of elemetns nums[i] is divided such that every number is
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      // less than mxm and minimum is maximized
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      val = ceil(nums[i] / (double)mxm);
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      // no. of steps is val-1
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      res += (val - 1);
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      // the new maximized minimum value
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      val = nums[i] / val;
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      mxm = val;
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    }
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    return res;
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  }
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};