每日一九度之 题目1039：Zero-complexity Transposition
时间限制：1 秒
内存限制：32 兆
特殊判题：否
提交：3372
解决：1392
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.
</dd> </dl> <dl> <dt> 输入： </dt> <dd>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).
</dd> </dl> <dl> <dt> 输出： </dt> <dd>For each case, on the first line of the output file print the sequence in the reverse order.
</dd> </dl> <dl> <dt> 样例输入： </dt> <dd>5 -3 4 6 -8 9</dd> </dl> <dl> <dt> 样例输出： </dt> <dd>
9 -8 6 4 -3</dd> </dl>
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 = 10005; typedef long long ll; int n, m; ll a[maxn]; int main(){ while( ~scanf("%d", &n) ){ for(int i=0; i<n; i++){ scanf("%ld",&a[i]); } for(int i=n-1; i>=0; i--){ printf(i==0?"%ld\n":"%ld ",a[i]); } } return 0; }
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 = 10005; typedef long long ll; int n, m; ll a[maxn]; int main(){ while( ~scanf("%d", &n) ){ for(int i=0; i<n; i++){ scanf("%ld",&a[i]); } for(int i=0; i<n/2; i++){ ll temp = a[i]; a[i] = a[n-1-i]; a[n-1-i] = temp; } for(int i=0; i<n; i++){ printf(i==n-1?"%ld\n":"%ld ",a[i]); } } return 0; }
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