Question

C程序随机输入一个字符串，统计该串中每个数字字符(0，1，2，3，4，5，6，7， 8，9)出现的次

C程序随机输入一个字符串，统计该串中每个数字字符(0，1，2，3，4，5，6，7， 8，9)出现的次数。
考虑：
（1）用num［0］来统计字符0的个数，用num［1］来统计字符1的个数......用num［9］来统计字符9的个数。
（2）如何实现一个字符串的输入。

Answer 1

#include <stdio.h>
typedef struct
{
unsigned int num0;
unsigned int num1;
unsigned int num2;
unsigned int num3;
unsigned int num4;
unsigned int num5;
unsigned int num6;
unsigned int num7;
unsigned int num8;
unsigned int num9;
}NUMCOUNT;

NUMCOUNT NumCount;
int main(void)
{
char buf[100];
memset(buf,0,100);
mmeset(&NumCount.num0,0,sizeof(NumCount));
printf("请输入一个字符串:\n");
scanf("%s",buf);
for(int i=0;i<strlen(buf);i++)
{
switch(buf[i]){
case 0:
NumCount.num0++;
break;
case 1:
NumCount.num0++;
break;
case 2:
NumCount.num0++;
break;
case 3:
NumCount.num0++;
break;
case 4:
NumCount.num0++;
break;
case 5:
NumCount.num5++;
break;
case 6:
NumCount.num6++;
break;
case 7:
NumCount.num7++;
break;
case 8:
NumCount.num8++;
break;
case 9:
NumCount.num9++;
break;
default:
break;
}
}

printf("0的个数：%d\n",NumCount.num0);
printf("1的个数：%d\n",NumCount.num1);
printf("2的个数：%d\n",NumCount.num2);
printf("3的个数：%d\n",NumCount.num3);
printf("4的个数：%d\n",NumCount.num4);
printf("5的个数：%d\n",NumCount.num5);
printf("6的个数：%d\n",NumCount.num6);
printf("7的个数：%d\n",NumCount.num7);
printf("8的个数：%d\n",NumCount.num8);
printf("9的个数：%d\n",NumCount.num9);

}