日期加减算法题【牛客平台】

KY258 日期累加.

#include
using namespace std;

const int N = 13;
int arr[N] = { 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int GetMonthDay(int year, int month)
{
    //判断闰年的情况
    if(month == 2 && (year % 4 == 0 && year % 100 != 0 ||  year % 400 == 0)) 
        return 29;
    //平常年
    return arr[month]; 
}

int main()
{
    int n = 0;
    cin >> n;
  
    
    for(int i = 0; i < n; i++)
    {
        int y , m ,d, num;
        cin >> y >> m >> d >> num;
        d += num;
         while(d > GetMonthDay(y, m))
         {
             d -= GetMonthDay(y, m);
             m++;
             if(m == 13) m = 1, y++;
         }
    
          printf("%4d-%02d-%02dn", y, m, d);
    }
   
    return 0;
}

KY222 打印日期.

#include
using namespace std;
const int N = 13;
int arr[N] = {0,31,28,31,30,31,30,31,31,30,31,30,31};
int  main()
{
    int y, m;
    while(cin >> y >> m)
    {
        if((y % 4 == 0 && y % 100 != 0) || y % 400 == 0) arr[2] += 1;  //润年
        int i = 0;
        for( i = 1; i <= 12; i++ )
        {
              if(m > arr[i]) m -= arr[i];  //当前天减去当前月份天数
              else break;  //相等或者小于的情况
        }
        printf("%4d-%02d-%02dn", y,i,m) ;  //大于年、月、天
    }
    
    return 0;
}

计算日期到天数转换.

#include
using namespace std;
const int N = 13;
int arr[N] = {0,31,28,31,30,31,30,31,31,30,31,30,31};

int main()
{
    int y, m, d; 
    while(cin >> y >> m >> d)
    {
        int ret = 0;
        if((y % 4 == 0 && y % 100 != 0) || y % 400 == 0) arr[2] += 1;
        for(int i = 1; m <= 12 && i <= m - 1; i ++)
        {
              ret += arr[i];
        }
        ret += d;
        printf("%dn",ret);
    }
    
    return 0;    
}