循环结构程序设计

输入一行字符，要求统计出英文字母、数字和其它字符的个数。

小弟最近看到一道C++题,百思不得其解,希望高人帮在下指点一二,不胜感激.

“输入一行字符” 不解，还是统计个数不解呢？

[CODE][/CODE]
string str ;
cout << "please type in the string" << endl ;
cin >> str ;
// define three counter for number, letter(upper,lower), other characters , separately.
int num = 0 ;
int letter = 0 ;
int otherChar = 0 ;
for(string::iterator iter = str.begin() ; iter != str.end() ; ++iter)
{
if ( (*iter == '0')
||(*iter == '1')
||(*iter == '2')
||(*iter == '3')
||(*iter == '4')
||(*iter == '5')
||(*iter == '6')
||(*iter == '7')
||(*iter == '8')
||(*iter == '9')
)
{
++num;
}
else if( (*iter == 'a')
||(*iter == 'A')
||(*iter == 'b')
||(*iter == 'B')
||(*iter == 'c')
||(*iter == 'C')
||(*iter == 'd')
||(*iter == 'D')
||(*iter == 'e')
||(*iter == 'E')
||(*iter == 'f')
||(*iter == 'F')
||(*iter == 'g')
||(*iter == 'G')
||(*iter == 'h')
||(*iter == 'H')
||(*iter == 'i')
||(*iter == 'I')
||(*iter == 'j')
||(*iter == 'J')
||(*iter == 'k')
||(*iter == 'K')
||(*iter == 'l')
||(*iter == 'L')
||(*iter == 'm')
||(*iter == 'M')
||(*iter == 'n')
||(*iter == 'N')
||(*iter == 'o')
||(*iter == 'O')
||(*iter == 'p')
||(*iter == 'P')
||(*iter == 'q')
||(*iter == 'Q')
||(*iter == 'r')
||(*iter == 'R')
||(*iter == 's')
||(*iter == 'S')
||(*iter == 't')
||(*iter == 'T')
||(*iter == 'u')
||(*iter == 'U')
||(*iter == 'v')
||(*iter == 'V')
||(*iter == 'w')
||(*iter == 'W')
||(*iter == 'x')
||(*iter == 'X')
||(*iter == 'y')
||(*iter == 'Y')
||(*iter == 'z')
||(*iter == 'Z'))
{
++letter ;
}
else
{
++otherChar ;
}
}
cout << "there is " << num << " numbers, " << letter << " letters, " << otherChar
<< " other characters. " << endl ;

应该是这样.

   难道你就不觉得麻烦吗?简化一下吧,很简单的.

isdigit()
isalpha()....之类的函数就可以了,自己找下~

4楼和6楼的方法都有些“炮打蚊子”的意思了
比较简单的
int main(){
unsigned int alpha, digit, other; //分别保存字母、数字、其他的字符数
unsigned char str[256]={0};
//读取字符串，假设已经放到str[]中(略)
long i=0;
while(str[i])
if((str[i]>='a'&&str[i]<='z')||(str[i]>='A'&&str[i]<='Z'))
++alpha;
elseif(str[i]>='0'&&str[i]<='9')
++digit;
else
++other;
return 0;
}

计算结果alpha即是字母数,digit是数字数,other是特殊字符数

LZ的题目有歧意。

恩~~~  是每个字符或数字的个数  还是如13楼所解？？~~~

7楼的方法很好，不过i貌似没有自增。。。，另外lz请说出不解的地方。

修改了一下：
#include<stdio.h>
void main()
{
int alpha=0, digit=0, other=0;
char str[256];
gets(str);
int i=0;
while(str[i])
{
if((str[i]>='a'&&str[i]<='z')||(str[i]>='A'&&str[i]<='Z'))
++alpha;
else
if(str[i]>='0'&&str[i]<='9')
++digit;
else
++other;
i++;
}
printf("%d\n%d\n%d\n",alpha,digit,other);
}

[此贴子已经被作者于2007-10-2 19:53:31编辑过]