題目鏈接：http://acm.hrbust.edu.cn/index.php?m=ProblemSet&a=showProblem&problem_id=1391

Orz odd
Time Limit: 1000 MS Memory Limit: 65536 K
Total Submit: 23(10 users) Total Accepted: 9(8 users) Rating: Special Judge: No
Description
An odd number n ≥ 3 is an “Orz odd” if and only if there exists a permutation of {1，2，…，n} such as {a1，a2，…，an} can make the following n formulas

a1 – a2 + a3 – a4 + … – an – 1 + an

a2 – a3 + a4 – a5 + … – an + a1

……

an – a1 + a2 – a3 + … – an – 2 + an – 1

are all positive.

Please determine if a given number n is an “Orz odd”.

Input
There are multiple test cases. The first line of input is an integer T indicating the number of test cases. Then T test cases follow.

For each test case:

Line 1. This line contains the single integer n (3?≤ n ≤?109).

Output
For each test case:

Line 1. Output “Orz” if n is an “Orz odd”; otherwise output “JiaozhuV5” instead.

Sample Input
2

4

5

Sample Output
JiaozhuV5

Orz

Hint
In the second sample, a permutation of {1，2，3，4，5} is {2，1，3，5，4}.

It can make the following 5 formulas

2 – 1 + 3 – 5 + 4 > 0

1 – 3 + 5 – 4 + 2 > 0

3 – 5 + 4 – 2 + 1 > 0

5 – 4 + 2 – 1 + 3 > 0

4 – 2 + 1 – 3 + 5 > 0

are all positive. So, 5 is an “Orz odd”.

Source
哈理工2012春季校賽 - 現場賽
Author
齊達拉圖@HRBUST

【思路分析】這個題想了很久，證明沒有證明出來為什么，大致的算證出來了，但是寫起來有點麻煩，用STL的next_permiation模擬了一下結果是：n%4=1的時候能使所有式子的結果是整數。
【AC代碼】

#include<cstdio> #include<cstring> #include<algorithm> using namespace std;int main() {int t,n;scanf("%d",&t);while(t--){scanf("%d",&n);if(n%4==1)//注意取余只要大于0都為邏輯真{printf("Orz\n");}else{printf("JiaozhuV5\n");}}return 0; }

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