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Contest100000578 《算法笔记》3.4小节——入门模拟->日期处理
1928-ProblemA-日期差值
来自 http://codeup.cn/contest.php?cid=100000578
题析：经典的日期模拟，注意平年闰年的差别，两个时间点之间的差值用较小的时间不断累加来追上较大的时间，累加时注意到达界限时上一个级别的更新。
//1928ProblemA日期差值
#include <iostream>
#include <stdlib.h>
#include <cstring>
using namespace std;
int monthDay[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}
};
bool isLeap(int year)
{
if((year%4==0&&year%100!=0) || year%400==0)
{
return true;
}
return false;
}
int main()
{
int time1,year1,month1,day1;
int time2,year2,month2,day2;
while(scanf("%d%d",&time1, &time2) != EOF)
{
if(time1>time2)
{
int temp=time1;
time1=time2;
time2=temp;
}
year1 = time1/10000;
month1 = time1%10000/100;
day1 = time1%100;
year2 = time2/10000;
month2 = time2%10000/100;
day2 = time2%100;
int count=0;//计数值
while(year1<year2 || month1<month2 || day1<day2)//较小的时间点累加追较大的时间点
{
day1++;
if(day1 > monthDay[month1][isLeap(year1)])//天数满月加一
{
month1++;
day1=1;
}
if(month1 > 12)//月数满年加一
{
year1++;
month1=1;
}
count++;
}
count=count+1;//如果两个日期是连续的我们规定他们之间的天数为两天
printf("%d\n",count);
}
return 0;
}
1929-ProblemB-Day of Week
来自 http://codeup.cn/contest.php?cid=100000578
题析：主要涉及基姆拉尔森计算公式，另外还需要注意：
字符数组的初始化问题
字符串比较函数strcmp==0时表示相等
//1929ProblemBDay of Week
#include <iostream>
#include <cstdlib>
#include <cstring>
using namespace std;
char monthName[13][20] = {{},
{"January"},{"February"},{"March"},{"April"},{"May"},
{"June"},{"July"},{"August"},{"September"},{"October"},
{"November"},{"December"}
};//月份
char weekday[7][20]={{"Monday"},{"Tuesday"},{"Wednesday"},{"Thursday"},{"Friday"},
{"Saturday"},{"Sunday"}
};//星期
/*
//基姆拉尔森计算公式外文名是Kim larsson calculation formula。
// W= (d+2*m+3*(m+1)/5+y+y/4-y/100+y/400+1)%7 //C++计算公式
在公式中d表示日期中的日数，m表示月份数，y表示年数。
注意：在公式中有个与其他公式不同的地方：
把一月和二月看成是上一年的十三月和十四月，
例：如果是2004-1-10则换算成：2003-13-10来代入公式计算。
*/
int main()
{
int W,d,m,y;
int day;
char month[15];
int year;
while(scanf("%d%s%d",&day, month, &year) != EOF)
{
d=day;
y=year;
for(int i=1;i<=12;i++)
{
if(strcmp(monthName[i],month)==0)//strcmp字符串相等时值为0
{
m=i;
}
}
if(m==1 || m==2)//注意此处 把一月和二月看成是上一年的十三月和十四月，
{
m+=12;
y--;
}
W=(d+2*m+3*(m+1)/5+y+y/4-y/100+y/400)%7;
printf("%s\n",weekday[W]);
}
return 0;
}
1931-ProblemC-打印日期
来自 http://codeup.cn/contest.php?cid=100000578
题析：日期问题的变形，需注意：
Num--与day++的同步，此处不会出现year++的情况
打印日期的格式，需scanf的格式化输出
但法一为什么50%错误率还需进一步探讨
//1931ProblemC打印日期
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int monthDay[13][10]={
{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}
};
bool isLeap(int year)
{
return (year%4==0 && year%100!=0 || year%400==0);
}
int main()
{
int year;
int num;
while(scanf("%d%d",&year, &num) != EOF)//输入参数
{
int day=0;//天数计数
int month=1;//月份计数
//法二
while(num>0)
{
num--;
day++;
if(day > monthDay[month][isLeap(year)])
{
day=1;
month++;
}
}
printf("%04d-%02d-%02d\n",year,month,day);
/*
//法一
for(i=1;i<12;i++)
{
if(day <= num)
day += monthDay[i][isLeap(year)];
else
break;
}
if(day<num)
{
day=num-day;
printf("%04d-%02d-%02d\n",year,12,day);
}
else
{
i--;
day=num + monthDay[i][isLeap(year)]-day;
//打印输出年月日
printf("%04d-%02d-%02d\n",year,i,day);
}
*/
}
return 0;
}
2026-ProblemD-日期类
来自 http://codeup.cn/contest.php?cid=100000578
题析：简单的日期类问题，需注意：
多点测试n--
输出格式%04%02%02
//2026ProblemD日期类
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int monthDay[13][10]={
{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}
};
bool isLeap(int year)
{
return (year%4==0&&year%100!=0 || year%400==0);
}
int main()
{
int n;
scanf("%d",&n);
int year,month,day;
while(n--)
{
scanf("%d%d%d",&year, &month,&day);
day++;
if(day>monthDay[month][isLeap(year)])
{
day=1;
month++;
}
if(month>12)
{
month=1;
year++;
}
printf("%04d-%02d-%02d\n",year,month,day);
}
return 0;
}
2063-ProblemE-日期累加
来自 http://codeup.cn/contest.php?cid=100000578
题析：
多点测试，n--
Num--与day++（涉及进位month++和year++）同步
//2063ProblemE日期累加
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int monthDay[13][10]={{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}
};
bool isLeap(int year)
{
return (year%4==0&&year%100!=0 || year%400==0);
}
int main()
{
int year,month,day,num;
int n;
scanf("%d",&n);
while(n--)
{
scanf("%d%d%d%d",&year,&month,&day,&num);
while(num--)
{
day++;
if(day>monthDay[month][isLeap(year)])
{
day=1;
month++;
}
if(month>12)
{
month=1;
year++;
}
}
printf("%04d-%02d-%02d\n",year,month,day);
}
return 0;
}
