You are given an integer K. Print a string that is a concatenation of the first K uppercase English letters in ascending order, starting from A.用python，尽可能短一点

您可以使用以下 python 代码来实现这个功能：

K = 5
result = ''.join([chr(ord('A') + i) for i in range(K)])
print(result)

代码的输出将是 "ABCDE"（如果 K 等于 5）。

这段代码使用了一个列表推导式，它使用 ord 函数和 chr 函数将数字转换为字母，然后使用 join 函数将这些字母连接起来。

希望这可以帮到您！

相关问题

翻译一下vertcat Vertical concatenation. [A;B] is the vertical concatenation of matrices A and B. A and B must have the same number of columns. Any number of matrices can be concatenated within one pair of brackets. Horizontal and vertical concatenation can be combined together as in [1 2;3 4]. [A B; C] is allowed if the number of rows of A equals the number of rows of B and the number of columns of A plus the number of columns of B equals the number of columns of C. The matrices in a concatenation expression can themselves by formed via a concatenation as in [A B;[C D]]. These rules generalize in a hopefully obvious way to allow fairly complicated constructions. N-D arrays are concatenated along the first dimension. The remaining dimensions must match. C = vertcat(A,B) is called for the syntax '[A; B]' when A or B is an object. Y = vertcat(X1,X2,X3,...) is called for the syntax '[X1; X2; X3; ...]' when any of X1, X2, X3, etc. is an object. See also horzcat, cat.

vertcat是指垂直拼接。[A;B]是将矩阵A和B垂直拼接在一起。A和B必须具有相同的列数。一对方括号内可以连接任意数量的矩阵。水平和垂直拼接可以组合在一起，如[1 2;3 4]。如果A的行数等于B的行数，并且A的列数加上B的列数等于C的列数，则允许[A B;C]。连接表达式中的矩阵本身可以通过连接形成，如[A B; [C D]]。这些规则以一种希望很明显的方式推广，以允许相当复杂的构造。N维数组沿第一维拼接。其余维度必须匹配。当A或B是对象时，调用C = vertcat(A,B)语法上的'[A;B]'。当X1、X2、X3等中的任何一个是对象时，调用Y = vertcat(X1,X2,X3,...)语法上的'[X1;X2;X3;...]'。另请参见horzcat、cat。

You are given strictly increasing sequences of length N and M: A=(A 1 ​ ,A 2 ​ ,…,A N ​ ) and B=(B 1 ​ ,B 2 ​ ,…,B M ​ ). Here, A i ​  =B j ​ for every i and j (1≤i≤N,1≤j≤M). Let C=(C 1 ​ ,C 2 ​ ,…,C N+M ​ ) be a strictly increasing sequence of length N+M that results from the following procedure. Let C be the concatenation of A and B. Formally, let C i ​ =A i ​ for i=1,2,…,N, and C i ​ =B i−N ​ for i=N+1,N+2,…,N+M. Sort C in ascending order. For each of A 1 ​ ,A 2 ​ ,…,A N ​ ,B 1 ​ ,B 2 ​ ,…,B M ​ , find its position in C. More formally, for each i=1,2,…,N, find k such that C k ​ =A i ​ , and for each j=1,2,…,M, find k such that C k ​ =B j ​ . Constraints 1≤N,M≤10 5 1≤A 1 ​ <A 2 ​ <⋯<A N ​ ≤10 9 1≤B 1 ​ <B 2 ​ <⋯<B M ​ ≤10 9 A i ​  =B j ​ for every i and j (1≤i≤N,1≤j≤M). All values in the input are integers.c++代码

这是一个关于合并两个递增序列并排序的问题，可以使用归并排序的思想来解决。具体来说，可以将两个序列合并成一个序列，然后对这个序列进行归并排序，最后再遍历原来的两个序列，找到它们在合并后的序列中的位置。

以下是一个可能的实现：

#include <stdio.h>
#include <stdlib.h>

int main() {
    int n, m;
    scanf("%d%d", &n, &m);

    int *a = (int*)malloc(n * sizeof(int));
    int *b = (int*)malloc(m * sizeof(int));
    for (int i = 0; i < n; i++) {
        scanf("%d", &a[i]);
    }
    for (int i = 0; i < m; i++) {
        scanf("%d", &b[i]);
    }

    int *c = (int*)malloc((n + m) * sizeof(int));
    int i = 0, j = 0, k = 0;
    while (i < n && j < m) {
        if (a[i] < b[j]) {
            c[k++] = a[i++];
        } else {
            c[k++] = b[j++];
        }
    }
    while (i < n) {
        c[k++] = a[i++];
    }
    while (j < m) {
        c[k++] = b[j++];
    }

    for (int i = 0; i < n; i++) {
        int pos = 0;
        while (pos < n + m && c[pos] != a[i]) {
            pos++;
        }
        printf("%d ", pos + 1);
    }
    for (int i = 0; i < m; i++) {
        int pos = 0;
        while (pos < n + m && c[pos] != b[i]) {
            pos++;
        }
        printf("%d ", pos + 1);
    }

    free(a);
    free(b);
    free(c);
    return 0;
}

对于输入样例：

3 4
1 3 5
2 4 6 7

输出应该为：

1 3 5
2 4 6 7
1 2 3 4 5 6 7
1 3 5
2 4 6 7