zoj 3594 Sexagenary Cycle 农历 天干地支

Sexagenary Cycle

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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The Chinese sexagenary cycle, also known as the stems-and-branches, is a cycle of sixty terms used for recording days or years. Each term in the sexagenary cycle consists of two Chinese characters, the first representing a term from a cycle of ten known as the heavenly stems and the second from a cycle of twelve known as the earthly branches. The first term (Jiazi) combines the first heavenly stem (Jia) with the first earthly branch (Zi). The second (Yichou) combines the second stem with the second branch. This continues, generating a total of 60 different terms (the least common multiple of ten and twelve), after which the cycle repeats itself.

The ten heavenly stems are Jia, Yi, Bing, Ding, Wu, Ji, Geng, Xin, Ren and Gui. And the twelve earthly branches are Zi, Chou, Yin, Mao, Chen, Si, Wu, Wei, Shen, You, Xu and Hai. E.g. Xinhai Revolution occurred in 1911, the year of the Xinhai stem-branch in the sexagenary cycle. And this year, namely 2012, is the year of Renchen. Given a year in western year, could you find out which year it is in cyclic year?

Actually, the cyclic year normally changes on the Chinese Lunar New Year, but you can ignore this in this problem.

Input

There are multiple test cases. The first line of input is an integer T ≈ 1000 indicating the number of test cases.

Each test case contains a positive integer or a negative integer. A positive integer n indicates n AD (Anno Domini), while a negative integer n indicates -n BC (Before Christ). The absolute values of all integers are strictly less than 10000.

Output

For each test case, output a string -- the corresponding cyclic year.

Sample Input

2
1911
2012

Sample Output

Xinhai
Renchen

链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3594

注意下公元没有0年，所以公元1年的前一年是公元前1年。

所以这题如果年份是公元前，那么要先+1  在加一个60的很大的倍数。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <malloc.h>
#include <ctype.h>
#include <math.h>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#include <stack>
#include <queue>
#include <vector>
#include <deque>
#include <set>
#include <map>

 //Jia, Yi, Bing, Ding, Wu, Ji, Geng, Xin, Ren and Gui.  //10
 //Zi, Chou, Yin, Mao, Chen, Si, Wu, Wei, Shen, You, Xu and Hai//12

string qian[10]={"Gui","Jia","Yi", "Bing", "Ding", "Wu", "Ji", "Geng", "Xin", "Ren"  };
string hou[12]={ "hai","zi","chou", "yin", "mao", "chen", "si", "wu", "wei", "shen", "you", "xu"};
int main()
{
	int t;
	cin>>t;
	int n;
	while(t--)
	{
		 cin>>n; 
		 if(n<0)
			 n++;
		 n=n-1911+8; 
		// n=
		// n-=2; 
		 int n2=n+4;
		 n+=120000;
		 n2+=120000;
		 cout<<qian[n%10]<<hou[n2%12]<<endl;
	} 
	return 0;
}