C语言（简单的）编写程序输入任意一串字符统计其中大写字母，小写字母。数字及其他字符的个数 C语言编程 从键盘输入一个字符串，分别统计其中大写字母、小写...
c\u8bed\u8a00\u4ece\u952e\u76d8\u4efb\u610f\u8f93\u5165\u4e00\u4e32\u5b57\u7b26\uff0c\u5206\u522b\u7edf\u8ba1\u5176\u4e2d\u5927\u5199\u5b57\u6bcd\uff0c\u5c0f\u5199\u5b57\u6bcd\u4e2a\u6570\u5177\u4f53\u6211\u5c31\u6ca1\u6cd5\u5199\uff0c\u5fd8\u8bb0\u5dee\u4e0d\u591a\u4e86\uff0c\u4f60\u628a\u5b57\u7b26\u4e32\uff0c\u5f04\u5230\u6570\u7ec4\u91cc\uff0c\u7136\u540e\u6bd4\u8f83\u4e0b\uff0c\u5927\u5c0f\u5199\u6709\u4e00\u4e2a\u6bd4\u8f83\u7684\uff0c\u8fd9\u6837\u7edf\u8ba1\u4e0b\u5c31\u53ef\u4ee5\u4e86
#include
void main()
{
int countd=0,countx=0,countk=0,counts=0,countq=0;//\u5206\u522b\u7528\u6765\u5bf9\u5927\u5199\u5b57\u6bcd\u3001\u5c0f\u5199\u5b57\u6bcd\u3001\u7a7a\u683c\u3001\u6570\u5b57\u3001\u5176\u4ed6\u5b57\u7b26\u505a\u8ba1\u6570
char s[100],*p;
printf("\u8bf7\u8f93\u5165\u4e00\u4e2a\u5b57\u7b26\u4e32\uff1a");
int i=0;
while((s[i]=getchar())!='\n')i++;
p=&s[0];
while(*p!='\n')
{
if((*p>='A')&&(*p<='Z'))countd++;
else
if((*p>='a')&&(*p<='z'))countx++;
else
if(*p==' ')countk++;// \u2018\u2019\u4e2d\u662f\u4e00\u4e2a\u7a7a\u683c
else
if((*p>='0')&&(*p<='9'))counts++;
else
countq++;
p++;
}
printf("\u5927\u5199\u5b57\u6bcd\u7684\u4e2a\u6570\u4e3a\uff1a%d \u5c0f\u5199\u5b57\u6bcd\u7684\u4e2a\u6570\u4e3a\uff1a%d \u7a7a\u683c\u4e2a\u6570\u4e3a\uff1a%d \n",countd,countx,countk);
printf("\u6570\u5b57\u4e2a\u6570\u4e3a\uff1a%d \u5176\u4ed6\u5b57\u7b26\u4e2a\u6570\u4e3a%d\n",counts,countq);
}
程序可按照以下流程执行：
1、输入字符串
2、对于字符串中的每一个字符，判断其为何种类型，并将相应的累加计数器加1。对于大小写字母和数字，可通过一个范围（大于等于某值和小于等于某值）来判断
代码如下：
#include <string.h>
#include <stdio.h>
int main()
{
char str[1000];
int lowAlpha, upAlpha, num, other;
int i;
lowAlpha = upAlpha = num = other = 0;
for (i = 0; i < strlen(str); i++)
if (str[i] >= 'a' && str[i] <= 'z')
lowAlpha++;
else if (str[i] >= 'A' && str[i] <= 'Z')
upAlpha++;
else if (str[i] >= '0' && str[i] <= '9')
num++;
else
other++;
printf("Upper:%d, Lower:%d, Number:%d, Other:%d
", upAlpha, lowAlpha, num, other);
return 0;
}
#include<stdio.h>
#include<string.h>
#define n 20 //你可以改变n的值，但要保证n至少大于你想输入的字符数目一位
void main()
{
int i,BC,lc,m; //BC为大写的字母，lc为小写的字母
char a[n];
i=BC=lc=m=0;
printf("输入字符串，以#号结束:\t");
gets(a);
while(a[i]!='\0')
{
if(a[i]>='A'&&a[i]<='Z') BC++;
else
if(a[i]>='a'&&a[i]<='z') lc++;
else
m++;
i++;
}
printf("大写字母个数为：%d\n",BC);
printf("小写字母个数为：%d\n",lc);
printf("其他字符个数为：%d\n",m);
}
我调试通过了的
#include <stdio.h>
void count(char*);
int main()
{
char ch[100]={0};
scanf("%s", ch);
count(ch);
return 0;
}
void count(char* ch)
{
//分别记录大写，小写，数字的个数。
int big=0, small=0, character=0,qita = 0;
while (*ch)
{
if ((*ch>='A')&&(*ch<='Z'))
{
++big;
}
else if ((*ch>='a')&&(*ch<='z'))
{
++small;
}
else if ((*ch>='0')&&(*ch<='9'))
{
++character;
}
else
{
++qita;
}
++ch;
}
printf("大写字母的个数是：%d\n", big);
printf("小写字母的个数是：%d\n", small);
printf("数字的个数是：%d\n", character);
printf("其他字符的个数是：%d\n", qita);
}
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