摘要： 用分治法求n个元素集合S中的最大、最小元素（要求每次平分成2个子集）void maxmin(int A[], int &e_max, int &e_min ,int low, int high) { int mid, x1, y1, x2, y2; if (high - low <= 1) { //<=,可能为1个 if (A[high] > A[low]) { e_max = A[high]; e_min = A[low]; } else { ... 阅读全文