[acwing weekly rematch] 61st weekly 20220723

One 、 Summary of this week's race

• The third question is geometry , First, I pushed the formula for half a day on the draft paper , Go to the scene again python How to keep a few decimal places .
• In the end wa Half day .
• And the answer to the third question is not unique .

Two 、 4497. Divide candy

link : 4497. Divide candy

1. Title Description

2. Thought analysis

grading Easy.
Sign in problem , Just hard write .

3. Code implementation

import io
import os
import sys
from collections import deque

if os.getenv('LOCALTESTACWING'):
    sys.stdin = open('input.txt')

if __name__ == '__main__':
    T = int(input())
    for _ in range(T):
        a,b,c = map(int, input().split())
        print((a+b+c)//2)

3、 ... and 、4498. The pointer

link : 4498. The pointer

1. Title Description

2. Thought analysis

• At first, there was no good idea , Look at the data range 15, So directly dfs enumeration .

3. Code implementation

import io
import os
import sys
from collections import deque

if os.getenv('LOCALTESTACWING'):
    sys.stdin = open('input.txt')

if __name__ == '__main__':
    n = int(input())
    arr = []
    for _ in range(n):
        arr.append(int(input()))
    m = len(arr)

    def dfs(index, s):
        if index == m:
            if s % 360 == 0:
                return True
            return False
        if dfs(index + 1, s + arr[index]):
            return True
        return dfs(index + 1, s - arr[index])


    print('YES' if  dfs(0, 0) else 'NO')

Four 、4499. A circle

link : 4499. A circle

1. Title Description

2. Thought analysis

Geometry problems .

• Condition one means that your circle should be in a given circle (x1,y1,R) Inside .
• Condition 2 means that your circle cannot cover a given point (x2,y2), But you can touch （ Fix it ）.
• Condition three means that your circle should be as big as possible , In other words, your circle should expand vigorously against this point .

---

• Special judgment if the given point is not in the circle , Then the circle we can construct is the given circle , Directly adjust the decimal point output .
• If in a circle , And the center of the circle , Then our circle diameter is the radius of a given circle , Obviously, the position of the center of the circle can be taken casually by rolling around the given center of the circle , Unlimited possibilities , Here is the y Fix .
• The rest of the case requires a set of formulas , You can see the picture below me , The code directly sets the conclusion .

3. Code implementation

import io
import os
import sys
from collections import deque

if os.getenv('LOCALTESTACWING'):
    sys.stdin = open('input.txt')


def sovle(R, x1, y1, x2, y2):
    if (y1 - y2) ** 2 + (x1 - x2) ** 2 >= R ** 2:
        return print("%.6f" % (x1 * 1.0), "%.6f" % (y1 * 1.0), "%.6f" % (R * 1.0))

    if x1 == x2 and y1 == y2:
        return print("%.6f" % ((x1 + R/2) ), "%.6f" % (y1 * 1.0), "%.6f" % (R / 2))
    
    c = ((y1 - y2) ** 2 + (x1 - x2) ** 2) ** 0.5

    c2 = c + R
    y3 = (y2 - c2 * (y2 - y1) / c)
    x3 = (x2 - c2 * (x2 - x1) / c)
    y4 = (y3 + y2) / 2
    x4 = (x3 + x2) / 2
    r = c2 / 2
    print("%.6f" % x4, "%.6f" % y4, "%.6f" % r)


if __name__ == '__main__':
    R, x1, y1, x2, y2 = map(int, input().split())
    sovle(R, x1, y1, x2, y2)

6、 ... and 、 Reference link

• nothing