題目原文：（id=413)
A sequence of number is called arithmetic if it consists of at least three elements and if the difference between any two consecutive elements is the same.
For example, these are arithmetic sequence:
1, 3, 5, 7, 9 7, 7, 7, 7 3, -1, -5, -9
The following sequence is not arithmetic.
1, 1, 2, 5, 7
A zero-indexed array A consisting of N numbers is given. A slice of that array is any pair of integers (P, Q) such that 0 <= P < Q < N.
A slice (P, Q) of array A is called arithmetic if the sequence:
A[P], A[p + 1], ..., A[Q - 1], A[Q] is arithmetic. In particular, this means that P + 1 < Q.
The function should return the number of arithmetic slices in the array A.
Example:
A = [1, 2, 3, 4] return: 3, for 3 arithmetic slices in A: [1, 2, 3], [2, 3, 4] and [1, 2, 3, 4] itself.給定一個數列，找到其中爲等差數列的子串的個數。方法是從前往後對數列進行掃描，每次把當前元素加入先前考慮的數列中，判斷增加了多少爲等差數列的子串。顯然這些子串肯定包含最後一個元素，只要記錄上次迭代時倒數2個元素的差值d，再判斷現在倒數2個元素差值是否與記錄的數字d相同，是則增加的子串數爲之前差值爲d的子串數再加1.掃描一遍數字時間複雜度爲O(n)
int numberOfArithmeticSlices(vector<int>& A) {
int ans1, ans2;
ans1 = 0;
ans2 = 0;
int count=0, d;
if (A.size()>1) d=A[1]-A[0];
for (int i = 2;i < A.size();i++) {
if (d == A[i] - A[i - 1]) {
count = count+1;
}
else {
count = 0;
d = A[i] - A[i - 1];
}
if (i % 2)
ans1 = ans2 + count;
else
ans2 = ans1 + count;
}
return max(ans1, ans2);
}