《算法笔记》3.4小节——入门模拟->日期处理 问题 B: Day of Week
内容导读
互联网集市收集整理的这篇技术教程文章主要介绍了《算法笔记》3.4小节——入门模拟->日期处理 问题 B: Day of Week,小编现在分享给大家,供广大互联网技能从业者学习和参考。文章包含3459字,纯文字阅读大概需要5分钟。
内容图文
![《算法笔记》3.4小节——入门模拟->日期处理 问题 B: Day of Week](/upload/InfoBanner/zyjiaocheng/604/474fe5d4bcfb408c8105e6eae7b409d2.jpg)
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 Cop
y
21 December 2012
5 January 2013
样例输出 Copy
Friday
Saturday
代码
初出茅庐,请多指教!
#include<stdio.h> //today is 18 February 2021,it is Thursday.
#include<string.h>
char January[20]="January";
char February[20]="February";
char March[20]="March";
char April[20]="April";
char May[20]="May";
char June[20]="June";
char July[20]="July";
char August[20]="August";
char September[20]="September";
char October[20]="October";
char November[20]="November";
char December[20]="December";
int month[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}};
int isleap(int year){
if((year%4==0&&year%100!=0)||year%400==0) return 1;
else return 0;
}
int main(){
int d,m,y;
char M[20];
int dayofweek[1000]={0};
int num=0;
while(scanf("%d%s%d",&d,M,&y)!=EOF){
if(y<1000||y>3000){
printf("year error!\n");
return 0;
}
if(strcmp(M,January)==0) m=1;
else if(strcmp(M,February)==0) m=2;
else if(strcmp(M,March)==0) m=3;
else if(strcmp(M,April)==0) m=4;
else if(strcmp(M,May)==0) m=5;
else if(strcmp(M,June)==0) m=6;
else if(strcmp(M,July)==0) m=7;
else if(strcmp(M,August)==0) m=8;
else if(strcmp(M,September)==0) m=9;
else if(strcmp(M,October)==0) m=10;
else if(strcmp(M,November)==0) m=11;
else if(strcmp(M,December)==0) m=12;
else {
printf("month error!\n");
return 0;
}
if(d<1||d>month[m][isleap(y)]){
printf("day error!\n");
return 0;
}
//printf("%d %d %d\n",d,m,y);
int x=0;
if(y*10000+m*100+d<=20210218){
while(y<2021||m<2||d<18){
d++;
if(d==month[m][isleap(y)]+1){
m++;
d=1;
}
if(m==13){
y++;
m=1;
}
x++;
}
//printf("x=%d\n",x);
if(x%7==0) dayofweek[num]=4; //2/11 2/18
else if(x%7==6) dayofweek[num]=5; //2/12 5
else if(x%7==5) dayofweek[num]=6; //2/13 6
else if(x%7==4) dayofweek[num]=7; //2/14 7
else if(x%7==3) dayofweek[num]=1; //2/15 1
else if(x%7==2) dayofweek[num]=2; //2/16 2
else if(x%7==1) dayofweek[num]=3; //2/17 3
}
else if(y*10000+m*100+d>20210218){
while(y>2021||m>2||d>18){
d--;
if(d==0){
m--;
if(m==0){
y--;
m=12;
}
d=month[m][isleap(y)];
}
x++;
}
if(x%7==0) dayofweek[num]=4; //2/25 4
else if(x%7==6) dayofweek[num]=3; //2/24 3
else if(x%7==5) dayofweek[num]=2; //2/23 2
else if(x%7==4) dayofweek[num]=1; //2/22 1
else if(x%7==3) dayofweek[num]=7; //2/21 7
else if(x%7==2) dayofweek[num]=6; //2/20 6
else if(x%7==1) dayofweek[num]=5; //2/19 5
}
num++;
}
int i;
for(i=0;i<num;i++){
if(dayofweek[i]==4) printf("Thursday\n");
else if(dayofweek[i]==5) printf("Friday\n");
else if(dayofweek[i]==6) printf("Saturday\n");
else if(dayofweek[i]==7) printf("Sunday\n");
else if(dayofweek[i]==1) printf("Monday\n");
else if(dayofweek[i]==2) printf("Tuesday\n");
else if(dayofweek[i]==3) printf("Wednesday\n");
}
return 0;
}
内容总结
以上是互联网集市为您收集整理的《算法笔记》3.4小节——入门模拟->日期处理 问题 B: Day of Week全部内容,希望文章能够帮你解决《算法笔记》3.4小节——入门模拟->日期处理 问题 B: Day of Week所遇到的程序开发问题。 如果觉得互联网集市技术教程内容还不错,欢迎将互联网集市网站推荐给程序员好友。
内容备注
版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 gblab@vip.qq.com 举报,一经查实,本站将立刻删除。
内容手机端
扫描二维码推送至手机访问。