两种方法解决

解题思路:

注意事项:

参考代码:

解法一：用公司算（吉姆拉尔森）

#include<iostream>

#include<string>

using namespace std;

string week[] = {"Mon","Tue","Wen","Thu","Fir","Sat","Sun"};

int y,m,d;

int main(){

while(cin>>y>>m>>d){

if(m < 3){

m += 12;

y--;

}

int w = (y + y /4 + y / 400 - y / 100 + 2 * m + 3 * (m + 1)/5 + d) % 7;

cout<<week[w]<<endl;

}

return 0;

}

解法二：（日期一遍遍前移）

#include<iostream>

using namespace std;

int monthday[13] = {

0,31,28,31,30,31,30,31,31,30,31,30,31

};

bool is_leap(int y)

{

return y%4 == 0 && y%100 != 0 || y%400 == 0; 

}

int daysOfMonth(int year,int month)

{

if(month == 2){

return 28 + is_leap(year);

}

else return monthday[month];

}

int main()

{

int y,m,d;

cin >> y >> m >> d;

int year = 2011,month = 11,day = 11,

week = 5;

while(year != y || month != m || day != d)

{

day ++;

if(day > daysOfMonth(year,month)){

day = 1;

month ++;

}

if(month == 13){

month = 1;

year ++;

}

week ++;

if(week == 8){

week = 1;

}

}

cout << week << endl;

return 0;

}