时间限制：1 秒
内存限制：32 兆
特殊判题：否
提交：3372
解决：1392
- 题目描述：
-
You are given a sequence of integer numbers. Zero-complexity transposition of the sequence is the reverse of this sequence. Your task is to write a program that prints zero-complexity transposition of the given sequence.
- 输入：
-
For each case, the first line of the input file contains one integer n-length of the sequence (0 ＜ n ≤ 10 000). The second line contains n integers numbers-a1, a2, …, an (-1 000 000 000 000 000 ≤ ai ≤ 1 000 000 000 000 000).
- 输出：
-
For each case, on the first line of the output file print the sequence in the reverse order.
- 样例输入：
-
5
-3 4 6 -8 9
- 样例输出：
-
9 -8 6 4 -3
水题，但是因为是英文，出了一点小（不）问（认）题（识）。
本来认为是绝对值排序，但是后来仔细一看，原来就只是把数组反过来输出就好了。
看懂题目很重要。。是的，真的很重要！
这是不改变数组，直接反过来输出的。
//Asimple
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cctype>
#include <cstdlib>
#include <stack>
#include <cmath>
#include <map>
#include <string>
#include <queue>
#define INF 100000
using namespace std;
const int maxn = ;
typedef long long ll;
int n, m;
ll a[maxn]; int main(){
while( ~scanf("%d", &n) ){
for(int i=; i<n; i++){
scanf("%ld",&a[i]);
}
for(int i=n-; i>=; i--){
printf(i==?"%ld\n":"%ld ",a[i]);
}
}
return ;
}
做了一点小处理，将数组反了过来。
//Asimple
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cctype>
#include <cstdlib>
#include <stack>
#include <cmath>
#include <map>
#include <string>
#include <queue>
#define INF 100000
using namespace std;
const int maxn = ;
typedef long long ll;
int n, m;
ll a[maxn]; int main(){
while( ~scanf("%d", &n) ){
for(int i=; i<n; i++){
scanf("%ld",&a[i]);
}
for(int i=; i<n/; i++){
ll temp = a[i];
a[i] = a[n--i];
a[n--i] = temp;
}
for(int i=; i<n; i++){
printf(i==n-?"%ld\n":"%ld ",a[i]);
}
}
return ;
}