问题 B: Day of Week
来源:互联网 发布:网络语音对讲系统 编辑:程序博客网 时间:2024/05/29 08:15
题目描述
We now use the Gregorian style of dating in Russia. The leap years are years with number divisible by 4 but not divisible by 100, or divisible by 400.
For example, years 2004, 2180 and 2400 are leap. Years 2004, 2181 and 2300 are not leap.
Your task is to write a program which will compute the day of week corresponding to a given date in the nearest past or in the future using today’s agreement about dating.
输入
There is one single line contains the day number d, month name M and year number y(1000≤y≤3000). The month name is the corresponding English name starting from the capital letter.
输出
Output a single line with the English name of the day of week corresponding to the date, starting from the capital letter. All other letters must be in lower case.
样例输入
21 December 2012
5 January 2013
样例输出
Friday
Saturday
#include <stdio.h>#include <string.h>#define isleap(x) (x % 100 != 0 && x % 4 == 0)|| x % 400 == 0 ? 1 : 0 int dayofMonth[13][2] = {{0,0},{31,31},{28,29},{31,31},{30,30}, {31,31},{30,30},{31,31},{31,31},{30,30},{31,31},{30,30},{31,31}};char dayofWeek[7][20] = {"Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"};char monthofYear[13][20] = {"","January","February","March","April","May", "June","July","August","September","October","November","December"};int cordOfDate(int year1,int month1,int day1){ int year2 = 2017; int month2 = 11; int day2 = 8; int tempY,tempM,tempD; int cnt = 0; int num1 = year1*10000 + month1*100 + day1; int num2 = year2*10000 + month2*100 + day2; if(num1 > num2){ tempY = year1; tempM = month1; tempD = day1; year1 = year2; month1 = month2; day1 = day2; year2 = tempY; month2 = tempM; day2 = tempD; } while(year1 < year2 || month1 < month2 || day1 < day2){ day1++; if(day1 == dayofMonth[month1][isleap(year1)] + 1){ month1++; day1 = 1; } if(month1 == 13){ year1++; month1 = 1; } cnt++; } return cnt;}int main(){ int day,year; char month[20]; int num,n; while(scanf("%d %s %d",&day,month,&year) != EOF){ for(int i = 0;i < 14;i++){ if(strcmp(month,monthofYear[i]) == 0){ num = i; break; } } printf("%d\n",num); n = cordOfDate(year,num,day); printf("%d\n",n); puts(dayofWeek[(n+2) % 7]); } return 0;}
- 问题 B: Day of Week
- Day of Week
- OJ_1043 Day of Week
- Day of Week
- Day of Week
- jobdu 7 Day of Week
- 题目7:Day of Week
- 题目1043:Day of Week
- 题目1043:Day of Week
- 题目1043:Day of Week
- 题目1043:Day of Week
- RPG ILE Day of Week
- 交大1043 day of week
- 题目1043:Day of Week
- 题目1043:Day of Week
- 题目1043:Day of Week
- 题目1043:Day of Week
- 九度1043 Day of Week
- IDEA内存溢出解决办法
- AS3.0升级埋坑之路
- 回首Java——String、StringBuffer与StringBuilder之间区别
- 第10周项目1(3)-中序线索化二叉树的算法验证
- 当popupWindow内嵌套recyclerView布局过长的时候会被顶上去的问题
- 问题 B: Day of Week
- 树莓派3装ubuntu mate 16 开机自动登陆
- Java基础:浅谈Java中的对象和引用
- vue pc端引入高德地图
- Android插件化学习之路(五)之代理Activity
- Openstack中更新表结构:修改表的某个字段使之可以为空
- jupyter notebook 执行结果与pythonIDE执行不一样
- VMware突然出现虚拟机不能上网,但主机能够上网
- 做一个完整的Java Web项目需要掌握的技能