关于该问题简单描述:假设有n个人排成一个圈。从第一个人开始报数,数到第m个人的时候这个人从队列里出列。然后继续在环里数后面第m个人,让其出列直到所有人都出列。最后一个出列的是胜出者。下面用链表模拟n个同学手拉手围成一个圈。如果m为1的话,该游戏没有了意思,因为这样的话,第n个人一定是胜出者,所以排除这种情况。解决该问题,有很多方法,本方法用的是循环单链表。如有不当之处,请读者指正!
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#include<iostream>
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//#include<stdlib.h>
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using namespace std;
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struct Note {
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int data;
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struct Note *next;
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};
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Note *CreateNote() {
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Note *first;
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first = new Note; //first = (Note *)malloc(sizeof(Note));
- first ->data = NULL; //创建头结点,并且不存放任何值
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return first;
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}
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Note *InitNote(Note *first, int n) {
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Note *head, *p;
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head = first;
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p = NULL;
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cout << "同学开始座次:" << endl;
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for (int i = 1; i <= n; i++) { //利用尾插法,构造链表
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p = new Note;
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head->next = p;
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p->data = i;
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cout << p->data << "--> ";
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head = p;
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}
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p->next = first->next; //形成换
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return first; //返回头结点
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}
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void Search(Note *q, int m) {
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cout << "依次出列同学:";
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if (m == 1) { //如何查找间隔为1,则终止程序
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cout << "游戏太无聊!";
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exit(-1);
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}
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for (int i = 1; q != q->next; q = q->next, i++) {
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if (i == m) { //当i = m时,执行其中的语句,并初始化i = 1,至于原因,读者画图便可知
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i = 1;
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Note *n;
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n = q->next;
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cout << q->next->data << "--> ";
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q->next = n->next;
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delete n;
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}
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}
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cout << q->data;
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cout << endl;
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cout << "获胜的是:" << q->data << "号同学" << endl;
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}
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int main() {
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Note *p, *q;
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p = CreateNote();
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q = InitNote(p, 5);
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cout << endl;
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Search(q, 2);
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来自 “ ITPUB博客 ” ,链接:http://blog.itpub.net/29876893/viewspace-1815055/,如需转载,请注明出处,否则将追究法律责任。
转载于:http://blog.itpub.net/29876893/viewspace-1815055/