如果用一个循环数组q[0..m-1]表示队列时,该队列只有一个队列头指针front,不设队列尾指针rear，求这个队列中从队列投到队列尾的元素个数（包含队列头、队列尾）。

#include <iostream>
using namespace std;

//循环队列(少用一个空间)长度
#define M (8+1)

typedef struct node {
	int index;
	int nextIndex;
} Node;

Node* init(int front, int len) {
	//限制少用一个空间,没有限制少用一个下标,所以front>M-1
	if (front > M - 1 || len > M - 1) {
		return NULL;
	}
	Node* nodes = new Node[len];

	for (int i = 0, j = front; i < len; ++i, ++j) {
		nodes[i].index = j % M;
		nodes[i].nextIndex = (j + 1) % M;
	}
	return nodes;
}

void printDescription(Node* nodes, int len) {
	if (nodes) {
		for (int i = 0; i < len; ++i) {
			cout << "index: " << nodes[i].index << " nextIndex: "
					<< nodes[i].nextIndex << endl;
		}
		/*
		 *循环队列(少用一个空间)长度计算公式:length = (rear-front+1+M)%M;
		 *为了判断队列是否空/满情况
		 *空: front == rear
		 *满：front == (rear+1)%M
		 */
		int length = (nodes[len - 1].index - nodes[0].index + 1 + M) % M;
		cout << "length: " << length << endl;
	} else {
		cout << "参数错误!!!" << endl;
	}
}