我怎样才能在z之后再次打印abcd？ [英] How can I print after z again abcd like that?

问题描述

接受字符C和正整数N作为输入。该程序必须从C开始打印N个字符。



边界条件：

1< = N< = 100



输入格式：

第一行包含由空格分隔的C和N.



输出格式：

第一行包含N个字符。



示例输入/输出1：

输入：

a 4



输出：

abcd



示例输入/输出2：

输入：

z 5



输出：

zabcd



我尝试过：

Accept a character C and a positive integer N as input. The program must print N characters starting from C.

Boundary Condition(s):
1 <= N <= 100

Input Format:
The first line contains C and N separated by space(s).

Output Format:
The first line contains N characters.

Example Input/Output 1:
Input:
a 4

Output:
abcd

Example Input/Output 2:
Input:
z 5

Output:
zabcd

What I have tried:

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

int main()
{
    char a;
    int n,b=0;
    scanf("%c",&a);
    scanf("%d",&n);
    for( char i=a;b<n;i++){
        printf("%c",i);
        b=b+1;
    }
}

推荐答案

由于字母表中有26个字母，您可以使用％ 26 在z之后再次从头开始。

然而，字符a不是0，所以你需要先减去它然后重新添加它。



例如：

Since there are 26 letters in the alphabet, you can use % 26 to start from the beginning again after z.
The character 'a' is not 0 however, so you need to subtract it first and then later re-add it.

For example:

for (int i = 0; i < n; i++) {
    printf("%c", (a - 'a' + i) % 26 + 'a');
}

for (i=0; i<n; ++i)
{
  putchar(a);
  a = (a == 'z') ? 'a' : (a + 1);
}

在循环os中有2个变量基本上就是这个想法，但你的问题是你正在混合它们的用法。 br />
保持它们的使用分开，有1个变量来计算打印的字母，另一个变量用于跟踪要打印的字母。

Having 2 variables in the loop os basically the idea, but your problem is the you are mixing their usage.
Keep their usage separated, have 1 variable to count printed letters, and the other variable to keep track of the letter to print.

for( int i=0;i<n;i++){ // count here
    printf("%c",a);
    if (a<'z') // set next letter here
        a=a+1;
    else
        a='a';
}

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