3 solutions

  • 0
    @ 2022-11-27 23:57:44
    #include<iostream>
    using namespace std;
    
    int days[] = {0,31,28,31,30,31,30,31,31,30,31,30,31};
    
    int is_leap(int y){   //判断闰年
    	return y % 400 == 0 || y % 4 == 0 && y % 100 != 0;
    }
    
    int daysOfMonth(int y, int m){
    	if( m == 2)
    		return is_leap(y) + 28;
    	return days[m];
    }
    
    int main(){
    	int y = 2000, m = 1, d = 1, w = 6;   // 年,月,日,周几
    	int ans = 0; 
    	while(y != 2020 || m != 10 || d != 2){
    		if(d == 1 || w == 1)
    			ans += 2;
    		else 
    			ans++;
    		d++;
    		if(d > daysOfMonth(y, m))
    			d = 1, m++;
    		if(m > 12)
    			m = 1, y++;
    		w++;
    		if(w == 8)
    			w = 1;
    	}
    	cout << ans << endl;
    	return 0;
    }
    
    • 0
      @ 2022-4-7 22:05:25
      #include<bits/stdc++.h>
      using namespace std;
      int week=6,sum;
      int main(){
      	for(register int year=2000;year<=2020;year++){
      		int	a[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
      		if((year%4==0&&year%100!=0)||year%400==0)a[2]=29;//闰年特判
      		for(register int month=1;month<=12;month++)
      			for(register int day=1;day<=a[month];day++){
      				if(day==1||week==1)sum+=2;//月初或者周一
      				else sum++;
      				week=(week+1)%7;//星期数增加
      				if(year==2020 && month==10 && day==1){
      					cout<<sum;
      					return 0;
      				}
      			}
      	}
      }
      • 0
        @ 2022-4-7 11:42:18

        闰年是能被400整除或者能整除4但不能整除100的年,闰年2月份是29天,非闰年是28天

        #include <bits/stdc++.h>
        using namespace std;
        #define ll long long
        #define endl "\n"
        int yue[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
        int main(){
        	ll ans=0;
        	int zhou=5;
        	for(int i=2000;i<=2020;i++){
        		if(i%400==0||(i%4==0&&i%100!=0)){
        			yue[2]=29;
        		}
        		else yue[2]=28;
        		for(int j=1;j<=12;j++){
        			for(int k=1;k<=yue[j];k++){
        				zhou++;
        				zhou%=8;
        				if(zhou==0) zhou=1;
        				if(zhou==1||k==1) ans+=2;
        				else ans++;
        				if(i==2020&&j==10&&k==1){
        					cout<<ans;
        					return 0;
        				}
        			}
        		}
        	}
        	return 0;
        }
        • 1

        Information

        ID
        6481
        Time
        1000ms
        Memory
        256MiB
        Difficulty
        7
        Tags
        # Submissions
        57
        Accepted
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