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using namespace std; | using namespace std; | ||
int arr | int arr[1100] = { 0, }; | ||
int main(){ | int main(){ | ||
arr | arr[0] = 1; | ||
arr | arr[1] = 1; | ||
int n, ans; | int n, ans; | ||
cin>> n; | cin>> n; | ||
for(int i = 2; i<=n; i++){ | for(int i = 2; i<=n; i++){ | ||
arr | arr[i] = (arr[i-1] + arr[i-2] * 2) % 10007; | ||
} | } | ||
cout<<arr | cout<<arr[n]<<endl; | ||
} | } | ||
=== 2×n 타일링 === | === 2×n 타일링 === | ||
| Line 30: | Line 30: | ||
using namespace std; | using namespace std; | ||
int arr | int arr[1100] = { 0, }; | ||
int main(){ | int main(){ | ||
arr | arr[0] = 1; | ||
arr | arr[1] = 1; | ||
int n, ans; | int n, ans; | ||
cin>> n; | cin>> n; | ||
for(int i = 2; i<=n; i++){ | for(int i = 2; i<=n; i++){ | ||
arr | arr[i] = (arr[i-1] + arr[i-2]) % 10007; | ||
} | } | ||
cout<<arr | cout<<arr[n]<<endl; | ||
} | } | ||
== 박인서 == | == 박인서 == | ||
* 2xn 타일링 2 | |||
#include <iostream> | |||
int a[1001]; | |||
int main() | |||
{ | |||
int n; | |||
std::cin >> n; | |||
a[1] = 1; | |||
a[2] = 3; | |||
for (int i = 3; i <= n; i++) { | |||
a[i] = (a[i - 1] + a[i - 2] * 2) % 10007; | |||
} | |||
std::cout << a[n]; | |||
return 0; | |||
} | |||
* 2xn 타일링 | |||
#include <iostream> | |||
int a[1001]; | |||
int main() | |||
{ | |||
int n; | |||
std::cin >> n; | |||
a[1] = 1; | |||
a[2] = 2; | |||
for (int i = 3; i <= n; i++) { | |||
a[i] = (a[i - 1] + a[i - 2]) % 10007; | |||
} | |||
std::cout << a[n]; | |||
return 0; | |||
} | |||
== 곽정흠 == | == 곽정흠 == | ||
| Line 51: | Line 89: | ||
== 박인서 == | == 박인서 == | ||
* 전형적인 Dynamic Programming | |||
== 곽정흠 == | == 곽정흠 == | ||
Latest revision as of 14:46, 26 March 2026
오늘의 문제
참가자
- 15이원준
- 박인서
코드
15이원준
2×n 타일링 2
#include<iostream>
using namespace std;
int arr[1100] = { 0, };
int main(){
arr[0] = 1;
arr[1] = 1;
int n, ans;
cin>> n;
for(int i = 2; i<=n; i++){
arr[i] = (arr[i-1] + arr[i-2] * 2) % 10007;
}
cout<<arr[n]<<endl;
}
2×n 타일링
#include<iostream>
using namespace std;
int arr[1100] = { 0, };
int main(){
arr[0] = 1;
arr[1] = 1;
int n, ans;
cin>> n;
for(int i = 2; i<=n; i++){
arr[i] = (arr[i-1] + arr[i-2]) % 10007;
}
cout<<arr[n]<<endl;
}
박인서
- 2xn 타일링 2
#include <iostream>
int a[1001];
int main()
{
int n;
std::cin >> n;
a[1] = 1;
a[2] = 3;
for (int i = 3; i <= n; i++) {
a[i] = (a[i - 1] + a[i - 2] * 2) % 10007;
}
std::cout << a[n];
return 0;
}
- 2xn 타일링
#include <iostream>
int a[1001];
int main()
{
int n;
std::cin >> n;
a[1] = 1;
a[2] = 2;
for (int i = 3; i <= n; i++) {
a[i] = (a[i - 1] + a[i - 2]) % 10007;
}
std::cout << a[n];
return 0;
}
곽정흠
아이디어
15이원준
박인서
- 전형적인 Dynamic Programming