Description
一个长度为len(1<=len<=1000000)的顺序表,数据元素的类型为整型,将该表分成两半,前一半有m个元素,后一半有len-m个元素(1<=m<=len),设计一个时间复杂度为O(N)、空间复杂度为O(1)的算法,改变原来的顺序表,把顺序表中原来在前的m个元素放到表的后段,后len-m个元素放到表的前段。
注意:交换操作会有多次,每次交换都是在上次交换完成后的顺序表中进行。
Input
第一行输入整数len(1<=len<=1000000),表示顺序表元素的总数;
第二行输入len个整数,作为表里依次存放的数据元素;
第三行输入整数t(1<=t<=30),表示之后要完成t次交换,每次均是在上次交换完成后的顺序表基础上实现新的交换;
之后t行,每行输入一个整数m(1<=m<=len),代表本次交换要以上次交换完成后的顺序表为基础,实现前m个元素与后len-m个元素的交换;
Output
输出一共t行,每行依次输出本次交换完成后顺序表里所有元素。
Sample
Input
10 1 2 3 4 5 6 7 8 9 -1 3 2 3 5
Output
3 4 5 6 7 8 9 -1 1 2 6 7 8 9 -1 1 2 3 4 5 1 2 3 4 5 6 7 8 9 -1
Hint
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define max 100000
struct sequencelist{
int *element;
int length;
int maxsize;
};
void create(struct sequencelist *seqlist, int len){
int i;
for(i = 0; i < len; i++){
scanf("%d", &seqlist -> element[i]);
}
seqlist -> length = len;
}
void inverse(struct sequencelist *seqlist, int b, int e){
int i;
int j;
int t;
for(i = b, j = e; i < j; i++, j--){
t = seqlist -> element[i];
seqlist -> element[i] = seqlist -> element[j];
seqlist -> element[j] = t;
}
}
void display(struct sequencelist *seqlist){
int i;
for(i = 0; i < seqlist -> length - 1; i++){
printf("%d ", seqlist -> element[i]);
}
printf("%d\n", seqlist -> element[seqlist -> length - 1]);
}
int main(){
int len;
scanf("%d", &len);
struct sequencelist *seqlist1;
seqlist1 = (struct sequencelist *)malloc(sizeof(struct sequencelist));
seqlist1 -> element = (int *)malloc(max * sizeof(int));
seqlist1 -> length = 0;
seqlist1 -> maxsize = max;
create(seqlist1, len);
int t;
scanf("%d", &t);
int i;
int j;
//struct sequencelist *seqlist2;
//seqlist2 = (struct sequencelist *)malloc(sizeof(struct sequencelist));
//seqlist2 -> element = (int *)malloc(max * sizeof(int));
//seqlist2 -> length = len;
//seqlist2 -> maxsize = max;
for(i = 0; i < t; i++){
int m;
scanf("%d", &m);
inverse(seqlist1, 0, m - 1);
inverse(seqlist1, m, len - 1);
inverse(seqlist1, 0, len - 1);
display(seqlist1);
}
return 0;
}