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题目描述
n只猴子围坐成一个圈，按顺时针方向从1到n编号。然后从1号猴子开始沿顺时针方向从1开始报数，报到m的猴子出局，再从刚出局猴子的下一个位置重新开始报数，如此重复，直至剩下一个猴子，它就是大王。
输入
输入两个整数n和m,1<=m<=n<=100。
输出
输出猴王的编号
样例输入 Copy
8 3
样例输出 Copy
7
#include<cstdio>
#include<cstring>
#include<iostream>
#include<iomanip>
#include<cmath>
#include<cstdlib>
using namespace std;
typedef int status;
typedef int elemType;
typedef struct node{
elemType num;
struct node *next;
}LNode,*linklist;
int main(){
int length,king = 0;
cin >> length;
int k;
cin >> k;
linklist p,q,head;
if(length == 1){
king = 1;
}
p = (linklist)malloc(sizeof(node));
q = (linklist)malloc(sizeof(node));
head = p;
head -> num = 1;
q = head;
for(int i = 1; i < length; i ++){
p = (linklist)malloc(sizeof(node));//申请链表空间
p -> num = i + 1;
q -> next = p;//尾插法，通过p申请新的，然后通过q指向p相连
q = p;//使p变为q
}
q -> next = head;//尾头相连
p = head;//使p,q都指向head
for(int i = 1; i < length; i++){
for(int j = 1; j < k-1; j++){
p = p -> next;
}
q = p -> next;
p -> next = q -> next;
p = q -> next;
free(q);
}
king = p -> num;
free(p);
if(k == 1){
king = length;
}
cout << king << endl;
}
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