九度OJ 1039:Zero-complexity Transposition(逆置) (基础题)
- 题目描述:
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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:
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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:
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For each case, on the first line of the output file print the sequence in the reverse order.
- 样例输入:
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5 -3 4 6 -8 9
- 样例输出:
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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 ****************************************************************/
编程算法爱好者。