2022-02-14每日刷题打卡

#include 
#include 

using namespace std;

const int N = 1010;

int n, m;
char a[N], b[N];
int f[N][N];

int main()
{
    scanf("%d%s", &n, a + 1);
    scanf("%d%s", &m, b + 1);

    for (int i = 0; i <= m; i ++ ) f[0][i] = i;
    for (int i = 0; i <= n; i ++ ) f[i][0] = i;

    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= m; j ++ )
        {
            f[i][j] = min(f[i - 1][j] + 1, f[i][j - 1] + 1);
            if (a[i] == b[j]) f[i][j] = min(f[i][j], f[i - 1][j - 1]);
            else f[i][j] = min(f[i][j], f[i - 1][j - 1] + 1);
        }

    printf("%dn", f[n][m]);

    return 0;
}

#include 
#include 
#include 

using namespace std;

const int N = 15, M = 1010;

int n, m;
int f[N][N];
char str[M][N];

int edit_distance(char a[], char b[])
{
    int la = strlen(a + 1), lb = strlen(b + 1);

    for (int i = 0; i <= lb; i ++ ) f[0][i] = i;
    for (int i = 0; i <= la; i ++ ) f[i][0] = i;

    for (int i = 1; i <= la; i ++ )
        for (int j = 1; j <= lb; j ++ )
        {
            f[i][j] = min(f[i - 1][j] + 1, f[i][j - 1] + 1);
            f[i][j] = min(f[i][j], f[i - 1][j - 1] + (a[i] != b[j]));
        }

    return f[la][lb];
}

int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i ++ ) scanf("%s", str[i] + 1);

    while (m -- )
    {
        char s[N];
        int limit;
        scanf("%s%d", s + 1, &limit);

        int res = 0;
        for (int i = 0; i < n; i ++ )
            if (edit_distance(str[i], s) <= limit)
                res ++ ;

        printf("%dn", res);
    }

    return 0;
}

#include 
#include 
using namespace std;

constexpr size_t MAXN = 5e3;
int dp[MAXN][27];
int v;

signed main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    memset(dp, 0x80, sizeof dp), dp[0][0] = 0;
    cin >> v;
    for (int i = 1; i < MAXN; i++)
        for (int j = 1; j <= 26; j++)
            for (int k = 0; k < i; k++)
                dp[i][j] = max(dp[i][j], dp[k][j - 1] + k * (i - k));
    int i = 0, j = 0, same = 0, flag = 1;
    while (flag) {
        j = 1, i++;
        for (; j <= 26; j++) {
            if (dp[i][j] >= v) { flag = 0; break; }
        }
    }
    while (i) {
        if (j == 1)
            cout << (char)'a';
        else if (dp[i][j - 1] >= v)
            j--, cout << (char)(j + 'a' - 1), v -= i - 1, same = 1;
        else
            cout << (char)(j + 'a' - 1), v -= i - 1 - same, same++;
        i--;
    }
}