在c语言中,如何从键盘输入一串包含数字和字母的字符串,以连续的数字构成一个数,最多不超过六位,超好 C语言编程题:从键盘输入一串字符,统计其中的数字与字母个数并...

\u5728c\u8bed\u8a00\u4e2d,\u5982\u4f55\u4ece\u952e\u76d8\u8f93\u5165\u4e00\u4e32\u5305\u542b\u6570\u5b57\u548c\u5b57\u6bcd\u7684\u5b57\u7b26\u4e32,\u4ee5\u8fde\u7eed\u7684\u6570\u5b57\u6784\u6210\u4e00\u4e2a\u6574\u6570\uff0c\u8d85\u8fc7\u516d\u4e2a\u5219\u53e6\u5916\u7ec4\u6210

#include#include#include#define INI20#define INC 10char* input;//\u6307\u5411\u52a8\u6001\u6570\u7ec4int get_input();//\u7528\u4e8e\u63a5\u6536\u8f93\u5165\u503c\u7684\u51fd\u6570\uff0c\u4f7f\u7528\u52a8\u6001\u6570\u7ec4\u53ef\u4ee5\u4efb\u610f\u8f93\u5165\uff0c\u65e0\u5927\u5c0f\u9650\u5236void ini();//\u7cfb\u7edf\u521d\u59cb\u5316bool is_exit();//\u68c0\u6d4b\u662f\u5426\u9000\u51faint split();//\u5c06\u5b57\u7b26\u4e32\u5206\u89e3\u51fa6\u4f4d\u7684\u6570\u5b57bool is_number(char number); //\u5224\u65ad\u4e00\u4e2achar\u7c7b\u578b\u7684\u5143\u7d20\u662f\u5426\u4e3a\u6570\u5b57void delete_all();//\u5220\u9664\u52a8\u6001\u6570\u7ec4int main(){ini();while (1){get_input();if (is_exit())break;split();delete_all();}delete_all();return 0;}int get_input(){int limit = INI,cnt = 0;char ch;input = (char *)malloc(sizeof(char)*INI);while ((ch = getchar())!='\n'){if (cnt + 1 >= limit){//\u5982\u679c\u6570\u7ec4\u5927\u5c0f\u4e0d\u591f\uff0c\u90a3\u4e48\u91cd\u65b0\u7533\u8bf7\u66f4\u5927\u7684\u7a7a\u95f4char* tmp = input;limit += INC;input = (char *)malloc(sizeof(char)*limit);for (int i = 0; i = '0')return true;return false;}bool is_exit(){if (strcmp(input, "exit") == 0)return true;return false;}void ini(){printf("\u8bf7\u8f93\u5165\u5b57\u7b26\u4e32\uff1a(exit\u9000\u51fa)\n");}void delete_all(){free(input);input = NULL;}

#include
int main()
{
char a[100];
char b[24];
int s[100] = { 0 };//\u7528\u4e8e\u5b58\u50a8\u5b57\u7b26\u7684\u4e2a\u6570
gets(a);//\u8f93\u5165\u5b57\u7b26
//\u5f00\u59cb\u6bd4\u8f83
for (int x = 0; x < 24; x++)
{
int c = 0;//\u8bb0\u5f55\u6bcf\u4e2a\u5b57\u7b26\u4e2a\u6570
b[x] = x + 97;//\u4e3a\u4e86\u8ba9b[0]\u662fa\uff0cb[1]\u662fb\u4f9d\u6b21\u7c7b\u63a8
for (int i = 0; i < 100; i++)
{
if (b[x] == a[i])
{
++c;
s[x] = c;
}
}
if (s[x]>=1)//\u53ea\u8f93\u51fa\u8f93\u5165\u4e2d\u6709\u7684\u5b57\u6bcd \u7684\u4e2a\u6570
{
printf("%c %d\n", b[x], s[x]);
}
}
getchar();
return 0;
}

\u6269\u5c55\u8d44\u6599\uff1agets()\u7528\u6cd5\uff1a
gets()\u51fd\u6570\u7528\u6765\u4ece\u6807\u51c6\u8f93\u5165\u8bbe\u5907(\u952e\u76d8)\u8bfb\u53d6\u5b57\u7b26\u4e32\u76f4\u5230\u56de\u8f66\u7ed3\u675f,\u4f46\u56de\u8f66\u7b26
\u4e0d\u5c5e\u4e8e\u8fd9\u4e2a\u5b57\u7b26\u4e32\u3002\u5176\u8c03\u7528\u683c\u5f0f\u4e3a:
gets(s);
\u5176\u4e2ds\u4e3a\u5b57\u7b26\u4e32\u53d8\u91cf(\u5b57\u7b26\u4e32\u6570\u7ec4\u540d\u6216\u5b57\u7b26\u4e32\u6307\u9488)\u3002
gets(s)\u51fd\u6570\u4e0escanf("%s",s)\u76f8\u4f3c,\u4f46\u4e0d\u5b8c\u5168\u76f8\u540c,\u4f7f\u7528scanf("%s",s)
\u51fd\u6570\u8f93\u5165\u5b57\u7b26\u4e32\u65f6\u5b58\u5728\u4e00\u4e2a\u95ee\u9898,\u5c31\u662f\u5982\u679c\u8f93\u5165\u4e86\u7a7a\u683c\u4f1a\u8ba4\u4e3a\u8f93\u5165\u5b57\u7b26\u4e32\u7ed3\u675f,
\u7a7a\u683c\u540e\u7684\u5b57\u7b26\u5c06\u4f5c\u4e3a\u4e0b\u4e00\u4e2a\u8f93\u5165\u9879\u5904\u7406,\u4f46gets()\u51fd\u6570\u5c06\u63a5\u6536\u8f93\u5165\u7684\u6574\u4e2a\u5b57\u7b26
\u4e32\u76f4\u5230\u56de\u8f66\u4e3a\u6b62\u3002
\u793a\u4f8b\uff1a
#include<stdio.h>
int main()
{
char s[20];
printf("What's your name?\n");
gets(s);
printf("%s\n",s);
return 0;
}
\u5f53\u8f93\u5165a bde\u56de\u8f66\u7684\u65f6\u5019\uff0c\u4f1a\u8f93\u51faa bde\u3002

#include<stdio.h>
int main()
{
char c[200] = {'\0'};
scanf("%s",c);
int i;
int sum = 0;
char max = '\0';
for(i = 0; i < 200; i++)
{
char buf = c[i];
int j;
int k = 0;
for(j = i; j < 200; j++)
{
if(c[j] == '\0')
break;
if(buf == c[j])
k++;
}
if(k > sum)
{
sum = k;
max = buf;
}
}
printf("%c出现的最多,出现了%d次\n",max,sum);
return 0;
}

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