[coding test] Baekjoon_level 9_기본 수학 2-2
2021. 6. 5. 22:59ㆍPython/코딩 테스트
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▶ coding test_Baekjoon_level 9_기본 수학 2-2
- 백준: baekjoon
- 코딩 및 결과: github
JeongJaeyoung0/coding_test
코딩테스트. Contribute to JeongJaeyoung0/coding_test development by creating an account on GitHub.
github.com
- ^(caret): XOR
a | b | a xor b |
0 | 0 | 0 |
0 | 1 | 1 |
1 | 0 | 1 |
1 | 1 | 0 |
>>> a=10; b=0
>>> a^b
Out: 10
2021.06.05
# coding test_Baekjoon_level 9_기본 수학 2-2
# 9020 (골드바흐의 추측)-1
# 시간 초과
s=[];o=10000;p=[*range(o+1)]
for i in p:
if 1<i:p[i*i::i]=-(i+~o//i)*[0]
if 1<i:s.append(i)
t=int(input())
while t>0:
n=int(input())
l=[i for i in range(len(s)) if n-s[i]>1]
b=[[s[i],s[u]] for i in l for u in l if n-s[i]-s[u]==0]
print(*(b[(len(b)-1)//2]))
t-=1
# 9020 (골드바흐의 추측)-2
# 시간 초과
import math
n=5100
t=[True for i in range(n+1)]
for i in range(2,int(math.sqrt(n))+1):
if t[i]==True:
a=2
while i*a<=n:
t[i*a]=False
a+=1
s=[i for i in range(2,n+1) if t[i]]
t=int(input())
while t>0:
n=int(input())
l=[i for i in range(len(s)) if n-s[i]>1]
b=[[s[i],s[u]] for i in l for u in l if n-s[i]-s[u]==0]
print(*(b[(len(b)-1)//2]))
t-=1
# 9020 (골드바흐의 추측)-3
def p(n):
d=[True]*n
h=int(n**0.5)
for i in range(2,h+1):
if d[i]==True:
for j in range(i+i,n,i):
d[j]=False
return [i for i in range(2,n) if d[i]==True]
q=[True]*10000
h=int(10000**0.5)
for i in range(2,h+1):
if q[i]==True:
for j in range(i+i,10000,i):
q[j]=False
q[0]=False
q[1]=False
for j in range(int(input())):
a=int(input())
b=p(a+1)
b.reverse()
w=a
for k in b:
g=a-k
if q[g] == True:
m=k-g
if 0<=m and k>=g:
x=k
y=g
print(f'{y} {x}')
# 1085 (직사각형에서 탈출)
x,y,w,h=map(int,input().split())
print(min(x,w-x,y,h-y))
# 3009 (네 번째 점)-1
a=[]
exec("a+=list(map(int,input().split()));"*3)
if a[0]==a[2]:x=a[4]
elif a[0]==a[4]:x=a[2]
else:x=a[0]
if a[1]==a[3]:y=a[5]
elif a[1]==a[5]:y=a[3]
else:y=a[1]
print(f'{x} {y}')
# 3009 (네 번째 점)-2
x=y=0;exec("a,b=map(int,input().split());x^=a;y^=b;"*3)
print(x,y)
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