九度OJ 1039:Zero-complexity Transposition(逆置) (基础题)
2024-09-07 14:40:04
时间限制:1 秒
内存限制:32 兆
特殊判题:否
提交:3093
解决:1255
- 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
思路:
数组倒序。注意用long long类型存储。
代码:
#include <stdio.h> #define N 10000 int main(void)
{
int n, i;
long long a[N]; while (scanf("%d", &n) != EOF)
{
for(i=0; i<n; i++)
scanf("%lld", &a[i]); for(i=n-1; i>0; i--)
printf("%lld ", a[i]);
printf("%lld\n", a[i]);
} return 0;
}
/**************************************************************
Problem: 1039
User: liangrx06
Language: C
Result: Accepted
Time:0 ms
Memory:916 kb
****************************************************************/
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