任意给定一个字符串，字符串中包含除了空白符、换行符之外的的任意字符。你的任务是统计出现在该字符串中的各字母（即“A—Z”，“a—z”）的个数(区分大小写)。

输入格式:

一个长度不超过100的非空字符串。字符串中不会出现空白符、换行符。

输出格式:

字符串中出现的字母的统计信息，每个字母的统计信息占一行，按照字母的ASCII码的顺序输出。

输入样例:

AAAsdf&^%DF879as

输出样例:

注意单词“time”不论单复数，一律输出复数形式“times”

The character A has presented 3 times.
The character D has presented 1 times.
The character F has presented 1 times.
The character a has presented 1 times.
The character d has presented 1 times.
The character f has presented 1 times.
The character s has presented 2 times.

代码如下：

#include<stdio.h>
int main()
{
	int i;
	char ar[100] = { 0 };
	scanf("%s", ar);
//注意这两个数组的引入和使用
	int a[26] = { 0 };
	int A[26] = { 0 };
	for (i = 0; ar[i] != '\0'; i++)
	{
		if (ar[i] >= 'a' && ar[i] <= 'z')
			a[ar[i] - 'a']++;
		else if (ar[i] >= 'A' && ar[i] <= 'Z')
			A[ar[i] - 'A']++;
	}
	for (i = 0; i < 26; i++)
		if (A[i] != 0)
			printf("The character %c has presented %d times.\n", i + 'A', A[i]);
	for (i = 0; i < 26; i++)
		if (a[i] != 0)
			printf("The character %c has presented %d times.\n", i + 'a', a[i]);
	return 0;

}

这里大家要注意我统计字母个数的方法。

两个数组的引入！！