Algorithm
lc1588_所有奇数长度子数组的和
思路
感觉就是记录(各点 i 距离中心的的距离 + 1) / 2 * a[i]的和
对于偶数还是不太成熟…只对于奇数比较有用，这个规律也没有验证，推荐还是看题解的左右奇数个包含自己的个数值
https://leetcode-cn.com/problems/sum-of-all-odd-length-subarrays/solution/suo-you-qi-shu-chang-du-zi-shu-zu-de-he-yoaqu/
代码
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class Solution {
public:
int sumOddLengthSubarrays(vector<int>& arr) {
int sum = 0;
int n = arr.size();
for (int i = 0; i < n; i++) {
int leftCount = i, rightCount = n - i - 1;
int leftOdd = (leftCount + 1) / 2;
int rightOdd = (rightCount + 1) / 2;
int leftEven = leftCount / 2 + 1;
int rightEven = rightCount / 2 + 1;
sum += arr[i] * (leftOdd * rightOdd + leftEven * rightEven);
}
return sum;
}
};
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Review
https://docs.ceph.com/en/latest/radosgw/adminops/#get-usage
Tips
POST跟GET请求的区别
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git cm -s -m "signed-by test by add '-s'"
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markdown展示diff效果, 在代码块后面加上 diff