九度OJ 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:1508
解决:570
- 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
- 来源:
- 2007年上海交通大学计算机研究生机试真题
/********************************** 日期:2013-2-19* 作者:SJF0115* 题号: 九度OJ 题目1039:Zero-complexity Transposition* 来源:http://ac.jobdu.com/problem.php?pid=1039* 结果:AC* 来源:2007年上海交通大学计算机研究生机试真题* 总结:**********************************/#include<stdio.h>#include<string.h>#include<stdlib.h>int main(){ int i,n; while(scanf("%d",&n) != EOF){ long long int a[n]; for(i = 0;i < n;i++){ scanf("%lld",&a[i]);} for(i = n - 1;i >= 0;i--){ printf("%lld",a[i]); if(i == 0){printf("\n");} else{printf(" "); } } }return 0;}
第二种方法:字符串/********************************** 日期:2013-2-19* 作者:SJF0115* 题号: 九度OJ 题目1039:Zero-complexity Transposition* 来源:http://ac.jobdu.com/problem.php?pid=1039* 结果:AC* 来源:2007年上海交通大学计算机研究生机试真题* 总结:**********************************/#include <iostream>#include <cstring>using namespace std;int main(){string Num[10000];int N,i;while (cin>>N){for (i=0;i<N;i++){cin>>Num[i];}for (i=N-1;i>0;i--){cout<<Num[i]<<" ";}cout<<Num[0]<<endl;}return 0;}