3n+1数列问题
TimeLimit: 1000MS Memory Limit: 65536KB
SubmitStatistic
有一天小标遇到了经典的3n+1数链问题,他想知道3n+1数链的前k个数是多少。下面小标来给你介绍一下3n+1数链是什么,给定一个数n,如果n为偶数,那么下一个数n1 = n / 2;否则n1 = 3 * n + 1; 如果n1为偶数,那么下一个数n2 = n1 / 2;否则n2 = 3 * n1 + 1; 如果n2为偶数,那么下一个数n3 = n2 / 2;否则n3 = 3 * n2 + 1; .....小标最近刚刚学习了链表,他想把这两个知识结合一下,所以,他想按照下面的规定去做。①起始n为10,k为5,链表为空②判断n为偶数,他会往链表头部加一个5(即n/2),此时链表中序列为5,n变为5-> NULL③接下来n==5,判断n为奇数,他会往链表尾部加一个16(即3*n+1),此时链表中序列为5-> 16 -> NULL④接下来n==16,判断n为偶数,他会往链表头部加一个8(即n/2),此时链表中序列为8-> 5 -> 16 -> NULL⑤接下来n==8,判断n为偶数,他会往链表头部加一个4(即n/2),此时链表中序列为4 -> 8 - > 5 -> 16 -> NULL⑥接下来n==4,判断n为偶数,他会往链表头部加一个2(即n/2),此时链表中序列为2 -> 4 - > 8 - > 5 -> 16 -> NULL到此时,小标得到了前k个数,那么输出这个序列。Ps: 为了变得更容易理解,简单来说就是n为偶数,加在链表头部,n为奇数,加在链表尾部
Input
多组输入。对于每组数据,每行两个整数1 <= n , k <= 1000,含义如上
Output
输出链表序列
ExampleInput
10 5
ExampleOutput
2 4 8 5 16
Hint
Author
K#include<stdio.h>#include<string.h>#include<stdlib.h>#include<math.h>#include<iostream>#include<algorithm>#include<stack>#include<queue>//#include<dqeue>struct node{ int data; struct node*next;};int main(){ int n; while(~scanf("%d",&n)) { struct node *head; head = (struct node*)malloc(sizeof(struct node)); head->next = NULL; struct node *tail = head; for(int i=1;i<=n;i++) { struct node *p = (struct node*)malloc(sizeof(struct node)); p->next = NULL; scanf("%d",&p->data); tail->next = p; tail = p; } int t; scanf("%d",&t); while(t--) { int kk; scanf("%d",&kk); if(kk==1) { int num,f = 1; scanf("%d",&num); tail = head; while(tail->next) { if(tail->next->data>num) { struct node*u = (struct node*)malloc(sizeof(struct node)); u ->data = num; u->next = tail->next; tail ->next = u; f = 0; break; } else tail = tail->next; } if(f) { struct node*u=(struct node*)malloc(sizeof(struct node)); u->data = num; u->next =NULL; tail->next = u; } } else if(kk==2) { int f = 1; tail = head; while(tail->next) { if(f) f = 0; else printf(" "); printf("%d",tail->next->data); tail = tail->next; } printf("/n"); } } while(head) { tail = head; head = head->next; free(tail); } } return 0;}/***************************************************User name: jk160505徐红博Result: AcceptedTake time: 556msTake Memory: 3300KBSubmit time: 2017-01-14 11:21:54****************************************************/
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