Envío 6804
- Autor: danidiaztech
- Fecha: 2022-12-08 05:20:21 UTC (Hace alrededor de 2 meses)
Caso # |
Resultado |
Tiempo |
Memoria |
#1 |

Correcto
|
0.006 s
|
2 KBi |
#2 |

Correcto
|
0.005 s
|
1 KBi |
#3 |

Correcto
|
0.005 s
|
1 KBi |
#4 |

Correcto
|
0.004 s
|
1 KBi |
#5 |

Correcto
|
0.007 s
|
1 KBi |
#6 |

Correcto
|
0.005 s
|
1 KBi |
#7 |

Correcto
|
0.007 s
|
1 KBi |
#8 |

Correcto
|
0.005 s
|
1 KBi |
#9 |

Correcto
|
0.006 s
|
1 KBi |
#10 |

Correcto
|
0.007 s
|
1 KBi |
#11 |

Correcto
|
0.005 s
|
1 KBi |
#12 |

Correcto
|
0.005 s
|
1 KBi |
#13 |

Correcto
|
0.004 s
|
1 KBi |
#14 |

Correcto
|
0.004 s
|
1 KBi |
#15 |

Correcto
|
0.006 s
|
1 KBi |
#16 |

Correcto
|
0.005 s
|
1 KBi |
#17 |

Correcto
|
0.004 s
|
1 KBi |
#18 |

Correcto
|
0.005 s
|
1 KBi |
#19 |

Correcto
|
0.005 s
|
1 KBi |
#20 |

Correcto
|
0.005 s
|
1 KBi |
#21 |

Correcto
|
0.046 s
|
8 KBi |
#22 |

Correcto
|
0.044 s
|
8 KBi |
#23 |

Correcto
|
0.037 s
|
8 KBi |
#24 |

Correcto
|
0.041 s
|
7 KBi |
#25 |

Correcto
|
0.047 s
|
7 KBi |
#26 |

Correcto
|
0.048 s
|
7 KBi |
#27 |

Correcto
|
0.039 s
|
7 KBi |
Puntos totales: 100 / 100
Código
// Made by Daniel Diaz (@Danidiaztech)
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define fastInp cin.tie(0); cout.tie(0); ios_base::sync_with_stdio(0);
#define forn(i, n) for (int i = 0; i < n; i++) // for in range in python
#define FOR(i, a, b) for (int i = a; i < b; i++) // for in range in python
#define int long long int
#define double long double
#define pb push_back
#define ff first
#define ss second
#define mk make_pair
typedef pair<int, int> pii;
const int MAX = 1e6;
const int MIN = -MAX;
const int INF = LLONG_MAX;
const int MINF = LLONG_MIN;
const int MOD = 1e9 + 7;
int32_t main() {
fastInp;
#if LOCAL
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
string x, y;
cin >> x >> y;
int n = x.size();
int m = y.size();
int dp[n + 1][m + 1];
// Borders
for (int j = 0; j <= n ;j++){
dp[j][0] = j;
}
for (int i = 0; i <= m; i++){
dp[0][i]= i;
}
for (int i =1; i <= n; i++){
for (int j = 1; j <= m; j++){
dp[i][j] = min({
dp[i -1][j] + 1,
dp[i][j - 1] + 1,
dp[i -1][j - 1] + (x[i - 1] != y[j - 1])
});
}
}
// Start calculating dp
cout << dp[n][m] << endl;
return 0;
}