[Training] Blue Bridge Cup Day 1251

1251

[Blue Bridge Cup 2015 preliminary round] galaxy bomb

Many man-made floating Planet X "bomb" in the vast space of the galaxy in X, used as a landmark in the universe.
After each bomb explosion can be set for how many days.
For example: Alpha bomb placed January 1, 2015, timed to 15 days, then it exploded in January 16, 2015.
There is a beta bomb, a day of b c month placement, the timing is n days, you calculate the exact date of its explosion.
Input
there is an input a plurality of sets of data, each data input line, each line of input four positive integers a, b, c, n
input to ensure dates between 1000-01-01 2020-01-01, date and valid.
not more than 1000 n
output
Please fill in this date, in the format of yyyy-mm-dd i.e. two four year date February 2. For example: 2015-02-19
Please exactly as written. Other words or symbols can not appear.
Sample input the Copy
, 2015. 1. 1 15
, 2014 1000. 9. 11
sample output from the Copy
2015-01-16
2017-08-05
prompts
the subject has been adapted.

note

1. Sort things out with a pen and paper to draw a picture
2. Good function to create reusable functions
3. Leap years: four years a leap, leap two hundred years, four hundred years leap

algorithm

python v1.0:

1. By month treatment
2. Creating isLeap (), used to determine leap year
3. Create T (), the parameters, m, y to determine the number of days of the month

python v2.0、C++ v1.0：

1. Analog timepiece, plus one cycle time, the time complexity of O (n), n depending on
2. Note that in C ++ using printf ( "% 02d", d) to control the output

It can be noted that: the code longer, faster speed qwq, the first monthly reduction month, when n increases, the speed of the two algorithms are similar

answer

python v1.0:

def isLeapYear(year):#四年一闰，两百年不闰，四百年闰
    if (not year % 100 == 0) and (year % 4 == 0) or (year % 400 == 0):
        return True
 
def T(m, y):
    is31Day = [1,3,5,7,8,10,12]
    if m in is31Day:
        return 31
    elif m == 2:
        if isLeapYear(y):
            return 29
        else:
            return 28
    else:
        return 30
 
def main(data):
    y, m, d, n = map(int,data.split(' '))
    acc = 0
    acc = d + n
    while True:
        if T(m, y) == 31:
            if acc <= 31:
                d = acc
                break
            else:
                acc = acc - 31
                if m == 12:
                    y = y + 1
                    m = 1
                else:
                    m = m + 1
        if T(m, y) == 30:
            if acc <= 30:
                d = acc
                break
            else:
                m = m + 1
                acc = acc - 30
        if T(m, y) == 29:
            if acc <= 29:
                d = acc
                break
            else:
                m = m + 1
                acc = acc - 29
        if T(m, y) == 28:
            if acc <= 28:
                d = acc
                break
            else:
                m = m + 1
                acc = acc - 28
    #print(acc)
    print('{:}-{:0>2}-{:0>2}'.format(y,m,d))
while True:
    main(input())

python v2.0:

def isLeapYear(year):#四年一闰，两百年不闰，四百年闰
    if (not year % 100 == 0) and (year % 4 == 0) or (year % 400 == 0):
        return True
    else:
        return False
 
def main(data):
    D = [31,28,31,30,31,30,31,31,30,31,30,31]
    y, m, d, n = map(int,data.split(' '))
    if isLeapYear(y):
        D[1] = 29
    for i in range(n):
        d = d + 1
        if d > D[m-1]:
            d = 1
            m = m + 1
        if m > 12:
            m = 1
            y = y + 1
        if isLeapYear(y):
            D[1] = 29
        else:
            D[1] = 28   
    print('{:}-{:0>2}-{:0>2}'.format(y,m,d))
while True:
    main(input())

c++ v1.0:

#include<iostream>
using namespace std;
int f(int x){
    if(x%4==0&&x%100!=0||x%400==0)
        return 1;
    else return 0;
} 
int d[12]={31,28,31,30,31,30,31,31,30,31,30,31};
int main(){
    int a,b,c,n;
    while(cin>>a>>b>>c>>n){
        for(int i=1;i<=n;i++){
            if(f(a)) d[1]=29;
            else d[1]=28;
            if(++c>d[b-1]) {c=1;b++;}
            if(b>12){
                b=1;
                ++a; }
    }
    printf("%d-%02d-%02d\n",a,b,c);
    } 
    return 0;
}
/**************************************************************
    Problem: 1251
    User: yanshanbei
    Language: C++
    Result: 正确
    Time:0 ms
    Memory:2084 kb
****************************************************************/