OJ字符串练习(统计字母)_统计字符。jone最近玩起了字符游戏,规则是这样的-优快云博客

本文介绍了一种统计字符串中字母（包括大写和小写）出现次数的方法，通过使用C语言编程实现，针对输入的字符串，程序能够准确地计算并输出各字母的出现次数。

题目描述
问题描述：
任意给定一个字符串，字符串中包含除了空格、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)。
输入与输出要求：
输入一个长度不超过100的非空字符串。字符串中不会出现空格、换行符。输出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。
程序运行效果：
AAAsdf&^%DF879as↙
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.↙
注意单词“time”不论单复数，一律输出复数形式“times”。

#include<stdio.h>
#include<string.h>
int main()
{
 char str[1000];
 scanf("%s",str);
 int len=strlen(str);
 int count1[999]={0};
 int count2[999]={0};
 for(int i=0;i<len;i++)
 {
  if(str[i]>='A'&&str[i]<='Z')
  {
   count1[str[i]-'A']++;
  }
  else if(str[i]>='a'&&str[i]<='z')
  {
   count2[str[i]-'a']++;
  }
 }
 for(int i=0;i<=25;i++)
 {
  if(count1[i]!=0)
  {
   printf("The character %c has presented %d times.\n",(char)(i+'A'),count1[i]);
  }
 }
 for(int i=0;i<=25;i++)
 {
  if(count2[i]!=0)
  {
   printf("The character %c has presented %d times.\n",(char)(i+'a'),count2[i]);
  }
 }
 return 0;
}

不知为何OJ的运行时错误

#include<stdio.h>
#include<string.h>
int main()
{
 char ch[100], st;
 int a[100] = { 0 }, b[100] = { 0 };
 int count1 = 0, count2 = 0, count3 = 0, count4 = 0, i = 0, j;
 gets(ch);
 while (ch[i] != 0)
 {
  if (ch[i] >= 'a'&&ch[i] <= 'z')
  {
   count1 = ch[i] - 'a';
   a[count1]++;
   // count3++;
  }
  if (ch[i] >= 'A'&&ch[i] <= 'Z')
  {
   count2 = ch[i] - 'A';
   // count4++;
   b[count2]++;
  }
  i++;
 }
 for (j = 0; j <= 25; j++)
 {
  if (b[j] != 0)
  {
   printf("The character %c has presented %d times.\n", (char)('A' + j), b[j]);
  } 
 }
 for (j = 0; j <= 25; j++)
 {
  if (a[j] != 0)
  {
   printf("The character %c has presented %d times.\n", (char)('a' + j), a[j]);
  }
 }
 return 0;
}

升级简练版

#include<stdio.h>
int main()
{
	char a[999];
	gets(a);
	int countBig[26]={0};
	int countSmall[26]={0};
	for(int i=0;a[i]!='\0';i++)
	{
		if(a[i]>='A'&&a[i]<='Z'){
			countBig[a[i]-'A']++;
		}
		if(a[i]>='a'&&a[i]<='z'){
			countSmall[a[i]-'a']++;
		}
	}
	for(int i=0;i<26;i++){
		if(countBig[i]!=0){
			printf("The character %c has presented %d times.\n",'A'+i,countBig[i]);
		}
	}
	for(int i=0;i<26;i++){
		if(countSmall[i]!=0){
			printf("The character %c has presented %d times.\n",'a'+i,countSmall[i]);
		}
	}
	return 0;
}