<p>Given an integer array <code>nums</code> and two integers <code>lower</code> and <code>upper</code>, return <em>the number of range sums that lie in</em> <code>[lower, upper]</code> <em>inclusive</em>.</p>

<p>Range sum <code>S(i, j)</code> is defined as the sum of the elements in <code>nums</code> between indices <code>i</code> and <code>j</code> inclusive, where <code>i &lt;= j</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [-2,5,-1], lower = -2, upper = 2
<strong>Output:</strong> 3
<strong>Explanation:</strong> The three ranges are: [0,0], [2,2], and [0,2] and their respective sums are: -2, -1, 2.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [0], lower = 0, upper = 0
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
	<li><code>-10<sup>5</sup> &lt;= lower &lt;= upper &lt;= 10<sup>5</sup></code></li>
	<li>The answer is <strong>guaranteed</strong> to fit in a <strong>32-bit</strong> integer.</li>
</ul>

class Solution {
public:
    int countRangeSum(vector<int>& nums, int lower, int upper) {
        const int n = nums.size();
        vector<long long> tmp(n + 1);  // 修正：确保临时数组大小为 (right - left + 1)

        vector<long long> psum(n + 1, 0);  // 使用 long long 类型来避免溢出
        int count = 0;

        // 计算前缀和
        for (int i = 1; i <= n; i++) {
            psum[i] = psum[i - 1] + static_cast<long long>(nums[i - 1]);  // 确保加法不会溢出
        }

        function<void(int, int, int)> merge = [&](int left, int mid, int right) {
            int start = mid + 1, end = mid + 1;

            // 计算符合条件的区间数目
            for (int i = left; i <= mid; i++) {
                while (start <= right && psum[start] - psum[i] < lower) {
                    start++;
                }
                while (end <= right && psum[end] - psum[i] <= upper) {
                    end++;
                }
                count += end - start;
            }

            //vector<long long> tmp(right - left + 1);  // 修正：确保临时数组大小为 (right - left + 1)
            int i = left, j = mid + 1;
            int p = 0;

            // 合并两个有序的前缀和子数组
            while (i <= mid || j <= right) {
                if (j > right || (i <= mid && psum[i] <= psum[j])) {
                    tmp[p++] = psum[i++];
                } else {
                    tmp[p++] = psum[j++];
                }
            }

            // 修正：将合并后的数组拷贝回原数组，确保索引在合理范围内
            for (int k = 0; k < p; k++) {
                psum[left + k] = tmp[k];
            }
        };

        function<void(int, int)> msort = [&](int left, int right) {
            if (left >= right) {
                return;
            }

            int mid = left + (right - left) / 2;
            msort(left, mid);
            msort(mid + 1, right);
            merge(left, mid, right);
        };

        msort(0, n);
        return count;
    }
};

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