给定按升序排列的n个不同整数存于数组a[1:n]中，请设计O(log n)的算法找到下标i，1 ≤ i ≤ n，使得a[i] = i，如不存在这样的下标，则返回0。要求写出函数实现。函数原型如下：int find(int a[],int n); 正确的程序是（ ）
A、int find (int [] a, int n) { int left = 1; int right = n ; int mid = ?(left + right)/2?; while (left <= right) { if (a[mid]="=" mid) return mid;> mid) right = mid – 1; else left = mid + 1; } return 0; }
B、int find (int [] a, int n) { int left = 1; int right = n ; while (left < right) { int mid = ?(left + right)/2?; if (a[mid] == mid) return mid; if (a[mid] > mid) right = mid – 1; else left = mid + 1; } return 0; }
C、int find (int [] a, int n) { int left = 1; int right = n ; while (left <= right) { if (a[mid]="=" mid) return mid;> mid) right = mid – 1; else left = mid + 1; int mid = ?(left + right)/2?; } return 0; }
D、int find (int [] a, int n) { int left = 1; int right = n ; while (left <= right) { int mid="?(left" + 2?; if (a[mid]="=" mid) return mid;> mid) right = mid – 1; else left = mid + 1; } return 0; }
