Created
January 2, 2021 01:11
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a=0 | |
b=0 | |
c=0 | |
d=0 | |
e=0 | |
f=0 | |
t=0 | |
for i0 in range(5): | |
a += 1 | |
for i1 in range(5): | |
b += 1 | |
for i2 in range(5): | |
c += 1 | |
for i3 in range(5): | |
d += 1 | |
for i4 in range(5): | |
e += 1 | |
print("a=" + str(a) + " b=" + str(b) + " c=" + str(c) + " d=" + str(d) + " e=" + str(e)) | |
if a==b or a==c or a==d or a==e or b==c or c==d or d==e: | |
f += 1 | |
else: | |
t += 1 | |
e = 0 | |
d = 0 | |
e = 0 | |
c = 0 | |
d = 0 | |
e = 0 | |
b = 0 | |
c = 0 | |
d = 0 | |
e = 0 | |
print("t=" + str(t) + " f=" + str(f)) | |
print("전체 경우의 수: " + str(t+f) + "\n조건과 일치하는 경우의 수: " + str(t)) | |
print("답: " + str(t)) | |
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a=0 | |
b=0 | |
c=0 | |
d=0 | |
e=0 | |
f=0 | |
t=0 | |
for i0 in range(5): | |
a += 1 | |
for i1 in range(5): | |
b += 1 | |
for i2 in range(5): | |
c += 1 | |
for i3 in range(5): | |
d += 1 | |
for i4 in range(5): | |
e += 1 | |
print("a=" + str(a) + " b=" + str(b) + " c=" + str(c) + " d=" + str(d) + " e=" + str(e)) | |
if a==b or a==c or a==d or a==e or b==c or c==d or d==e: | |
f += 1 | |
else: | |
t += 1 | |
e = 0 | |
d = 0 | |
e = 0 | |
c = 0 | |
d = 0 | |
e = 0 | |
b = 0 | |
c = 0 | |
d = 0 | |
e = 0 | |
print("t=" + str(t) + " f=" + str(f)) | |
print("전체 경우의 수: " + str(t+f) + "\n조건과 일치하는 경우의 수: " + str(t)) | |
print("답: " + str(t)) | |
오오 파이썬을 여따가 쓰시다니.. 우왕 똑똑하시다,,
ㄷㄷ
속도가 느리니 C++로 다시 짜 봅시다
속도가 느리니 C++로 다시 짜 봅시다
#include <iostream>
using namespace std;
int main(){
int a = 0;
int b = 0;
int c = 0;
int d = 0;
int e = 0;
int f = 0;
int t = 0;
for (int i0=0; i0<5; i0++){
a += 1;
for (int i1=0; i1<5; i1++){
b += 1;
for (int i2=0; i2<5; i2++){
c += 1;
for (int i3=0; i3<5; i3++){
d += 1;
for (int i4=0; i4<5; i4++){
e += 1;
cout << "a=" << a << " b=" << b << " c=" << c << " d=" << d << " e=" << e << endl;
if (a==b or a==c or a==d or a==e or b==c or c==d or d==e){
f += 1;
}
else {
t += 1;
}
};
e = 0;
};
d = 0;
e = 0;
};
c = 0;
d = 0;
e = 0;
};
b = 0;
c = 0;
d = 0;
e = 0;
};
cout << "t=" << t << " f=" << f << endl;
cout << "전체 경우의 수: " << t+f << "\n조건과 일치하는 경우의 수: " << t << endl;
cout << "답: " << t << endl;
return 0;
};
짜왔습니다.
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와샌즈네
나도 저런거 만들어 보고 싶당