ZOJ 3594 Sexagenary Cycle

题意：天干地支。

        天干： Jia, Yi, Bing, Ding, Wu, Ji, Geng, Xin, Ren and Gui

        地支： Zi, Chou, Yin, Mao, Chen, Si, Wu, Wei, Shen, You, Xu and Hai

        每一轮是60次，不要误认为120次。。。（常识），这道题是经典的水题，很水但是还是被坑了很久大概一下午加一晚上。。。。。搞不懂自己什么水平

说下这道题目的坑把，首先是输出地支的时候是小写。。。其次还有AD和BC之分刚开始没有看见要求BC，最后是定义数组后当取余为0时候应该换到最后一个所以str[(t+9)%12]刚开始的时候把天干也当成12个了 害我找了很久

思路：刚开始看了维基百科里的解法，这也是第一次看英文会的解法，求AD，若n>=4则，n=（n-3）%60，若n=1-3，则代表58th，59th，60th。求BC，n=60-(n+2)%60.

         第二种方法简单，根据1911推导下就ok了，至于BC就是倒着来，和AD刚好相反

       

#include<stdio.h>
int main()
{
    char str1[11][10]={"Jia","Yi","Bing","Ding","Wu","Ji","Geng","Xin","Ren","Gui"};
    char str2[13][10]={"zi","chou","yin","mao","chen","si","wu","wei","shen","you","xu","hai"};
    long n,t1,t2;
    scanf("%ld",&n);
    while(n--)
    {
        int x;
        scanf("%ld",&x);
        if(x==1) printf("Xinyou\n");
        else if(x==2) printf("Renxu\n");
        else if(x==3) printf("Guihai\n");
        else if(x>=4)
        {
            x-=3;
            x=x%60; 
            t1=x%10;
            t2=x%12;
            printf("%s%s\n",str1[(t1+9)%10],str2[(t2+11)%12]);
        }
        else
        {
            x*=-1;
            x+=2;
            x=60-x%60;
            t1=x%10;
            t2=x%12;
            printf("%s%s\n",str1[(t1+9)%10],str2[(t2+11)%12]);
        }
    }
    return 0;
}

#include <stdio.h>
int main(int argc, char *argv[])
{
    int T,n;
    int t1,t2;
    char str1[15][10]={"Xin","Ren","Gui","Jia","Yi","Bing","Ding","Wu","Ji","Geng"};
    char str2[15][10]={"you","xu","hai","zi","chou","yin","mao","chen","si","wu","wei","shen"};
    char str3[11][10]={"Geng","Ji","Wu","Ding","Bing","Yi","Jia","Gui","Ren","Xin"};
    char str4[15][10]={"shen","wei","wu","si","chen","mao","yin","chou","zi","hai","xu","you"};
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        if(n>0)
        {
           n%=60;
           t1=n%10;
           t2=n%12;
           printf("%s%s\n",str1[(t1+9)%10],str2[(t2+11)%12]);
        }
        else
        {
            n*=-1;
            n%60;
            t1=n%10;
            t2=n%12;
            printf("%s%s\n",str3[(t1+9)%10],str4[(t2+11)%12]);
        }
    }
    return 0;
}