二分法插队

//by 温柔床乃英雄冢//谭浩强第二版习题5.4int a[] = {2,5,12,50,99,101,500};//已知顺序的序列,插入新值const int n = 7;int head = 0;int tear = n-1;int mid;int x, rank;cout << "input:";cin >> x;while (a[head] < x&&x < a[tear]) {//采用二分法提高时效if (tear - head <= 1)//防止进入死循环break;mid = (head + tear) / 2;if (a[mid] <= x) {head = mid;}else {tear = mid;}}if (x < a[head]) {rank = head;//大小差别1}else if (x>a[tear]) {rank = tear + 1;//3}else {rank = tear;//2}for (int i = 0; i < rank; i++) {//采用从rank处分割,"插入输出x"cout << a[i] << " ";}cout << x << " ";for (int i = rank; i < n; i++) {cout << a[i]<<" ";}