| 等 级 | |
| 排 名 | 1080 |
| 经 验 | 3126 |
| 参赛次数 | 2 |
| 文章发表 | 17 |
| 年 龄 | 21 |
| 在职情况 | 学生 |
| 学 校 | |
| 专 业 | 软件工程 |
自我简介：
TA的其他文章
你可能喜欢
母牛的故事 （C语言代码）浏览：983 |
人民币问题 （C语言代码）浏览：1465 |
Biggest Number (错误代码时间超限制)浏览：2105 |
不容易系列 （C语言代码）浏览：661 |
【求帮助】C语言程序设计教程（第三版）课后习题8.9 （C语言代码）浏览：792 |
解题思路:通过题意计算出第0年的纪年来作为初始值
注意事项:
参考代码:
#include<iostream>
using namespace std;
int main(){
string tiangan[10]={"jia","yi","bing","ding","wu","ji","geng","xin","ren","gui"};
string dizhi[12]={"zi","chou","yin","mao","chen","si","wu","wei","shen","you","xu","hai"};
int n,tstar=6,dstar=8;
cin>>n;
n%=60;
cout<<tiangan[(n%10+tstar)%10]<<dizhi[(n%12+dstar)%12];
return 0;
}0.0分
3 人评分
评论区
#include<iostream>
using namespace std;
string tian[10] = {"geng","xin","ren","gui","jia","yi","bing","ding","wu","ji"};
string di[12] = {"zi","chou","yin","mao","chen","si","wu","wei","shen","you","xu","hai"} ;
int main()
{
int n;
cin>>n;
if(n>0)
{
//2020年是庚子年
int d = n>2020?(n-2020):(2020-n);
int a = d%10;
int b = d%12;
cout<<tian[a]<<di[b];
}
return 0;
}
//可以看看我的
liuzhendong 2024-01-26 14:06:22 |
我是把庚子分别放到数组的第一个,然后对10和对12的余数作为偏移量,因为庚子是第一个所以这个就是下标
大佬们，请问怎么理解tstar=6和dstar=8呀？
笔盖大失 2022-04-01 23:49:02 |
天干初始在下标为6的地方即 庚 同理,初始地支在下标为8的地方即 申
叮叮咚咚嗒 2022-04-02 14:02:22 |
好的，谢谢
liuzhendong 2024-01-26 14:04:10 |
#include<iostream> using namespace std; string tian[10] = {"geng","xin","ren","gui","jia","yi","bing","ding","wu","ji"}; string di[12] = {"zi","chou","yin","mao","chen","si","wu","wei","shen","you","xu","hai"} ; int main() { int n; while(cin>>n) if(n>0) { //2020年是庚子年 int d = n>2020?(n-2020):(2020-n); int a = d%10; int b = d%12; cout<<tian[a]<<di[b]<<endl; } return 0; }
让我写了60行的代码情何以堪
liuzhendong 2024-01-26 14:03:56 |
#include<iostream> using namespace std; string tian[10] = {"geng","xin","ren","gui","jia","yi","bing","ding","wu","ji"}; string di[12] = {"zi","chou","yin","mao","chen","si","wu","wei","shen","you","xu","hai"} ; int main() { int n; while(cin>>n) if(n>0) { //2020年是庚子年 int d = n>2020?(n-2020):(2020-n); int a = d%10; int b = d%12; cout<<tian[a]<<di[b]<<endl; } return 0; }
- «
- 1
- »