标签:tst list dig The sub div begin end input
Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K=3, then you must output 3→2→1→6→5→4; if K=4, you must output 4→3→2→1→5→6.
Each input file contains one test case. For each case, the first line contains the address of the first node, a positive N (≤) which is the total number of nodes, and a positive K (≤) which is the length of the sublist to be reversed. The address of a node is a 5-digit nonnegative integer, and NULL is represented by -1.
Then N lines follow, each describes a node in the format:
Address Data Next
where Address
is the position of the node, Data
is an integer, and Next
is the position of the next node.
For each case, output the resulting ordered linked list. Each node occupies a line, and is printed in the same format as in the input.
00100 6 4
00000 4 99999
00100 1 12309
68237 6 -1
33218 3 00000
99999 5 68237
12309 2 33218
00000 4 33218 33218 3 12309 12309 2 00100 00100 1 99999 99999 5 68237 68237 6 -1
/* Name: Copyright: Author: 流照君 Date: 2019/8/5 14:52:14 Description: */ #include <iostream> #include<string> #include <algorithm> #include <vector> using namespace std; const int inf=100100; int main(int argc, char** argv) { //freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); int num=0,first,n,k,list[inf],add[inf],temp,next[inf]; cin>>first>>n>>k; for(int i=1;i<=n;i++) { cin>>temp; cin>>add[temp]>>next[temp]; } while(first!=-1) //不是所有输入都是有用的 { list[num++]=first; first=next[first]; } for(int i=0;i<num-num%k;i=i+k) reverse(begin(list)+i,begin(list)+i+k); //反转函数 for(int i=0;i<num-1;i++) printf("%05d %d %05d\n",list[i],add[list[i]],list[i+1]); printf("%05d %d -1", list[num - 1], add[list[num - 1]]); return 0; }
特别说一下
C++11引入了 begin 和 end 的函数,这两个函数与容器中的两个同名成员功能类似,不过这两个函数不是成员函数,而是含有参数的函数。
用法说明:
begin 返回首元素的地址,end 返回尾元素的下一个地址。
标签:tst list dig The sub div begin end input
原文地址:https://www.cnblogs.com/liuzhaojun/p/11398443.html