0001 void qWinSum(Queue<int>& A, Rank n, Rank w) { //w<=n
0002    qPreSum(A, n); //先求前缀和
0003    Queue<int> W; W.enqueue(0); //建立窗口
0004    for (Rank i = 1; i < w; i++) //充满窗口
0005       W.enqueue(A.dequeue());
0006    for (Rank i = w-1; i < n; i++) { //逐项
0007       A.enqueue(A.front() - W.dequeue()); //求差
0008       W.enqueue(A.dequeue()); //滑动窗口
0009    } //for
0010 } //qWinSum, O(n)