http://acm.hdu.edu.cn/showproblem.php?pid=2133

Problem Description
Today is Saturday, 17th Nov,2007. Now, if i tell you a date, can you tell me what day it is ?
 
Input
There are multiply cases.
One line is one case.
There are three integers, year(0<year<10000), month(0<=month<13), day(0<=day<32).
 
Output
Output one line.
if the date is illegal, you should output "illegal". Or, you should output what day it is.
 
Sample Input
2007 11 17
 
Sample Output
Saturday
 
时间复杂度:$O(1)$
代码:

#include <bits/stdc++.h>
using namespace std; int a[13] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int b[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
char s[8][10] = {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"};
bool IsRunNian(int year) {
if((year % 4 == 0 && year % 100 != 0) || year % 400 == 0)
return true;
return false;
}
int main() {
int year, month, day;
while(~scanf("%d%d%d", &year, &month, &day)) {
if(IsRunNian(year)) {
if(day > a[month] || month == 0 || day == 0) {
printf("illegal\n");
continue;
}
} else{
if(day > b[month] || month == 0 || day == 0) {
printf("illegal\n");
continue;
}
}
int sum = 0;
for(int i = 1; i < year; i ++) {
if(IsRunNian(i))
sum += 366;
else
sum += 365;
sum %= 7;
}
for(int i = 0; i < month; i ++) {
if(IsRunNian(year))
sum += a[i];
else
sum += b[i];
sum %= 7;
} sum += day;
sum %= 7;
printf("%s\n",s[sum]);
}
return 0;
}

  

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