c语言 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。 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...

c\u8bed\u8a00 \u968f\u673a\u8f93\u5165\u4e00\u4e2a\u5b57\u7b26\u4e32\uff0c\u7edf\u8ba1\u8be5\u4e32\u4e2d\u6bcf\u4e2a\u6570\u5b57\u5b57\u7b26(0\uff0c1\uff0c2\uff0c3\uff0c4\uff0c5\uff0c6\uff0c7\uff0c 8\uff0c9)\u51fa\u73b0\u7684\u6b21\u6570\u3002

\u601d\u8def\uff1a\u7edf\u8ba1\u5b57\u7b26\u4e32\u4e2d\u6bcf\u4e2a\u6570\u5b57\u5b57\u7b26\u51fa\u73b0\u7684\u4e2a\u6570\u53ef\u4ee5\u5148\u5b9a\u4e49\u4e00\u4e2a\u6574\u6570\u6570\u7ec4a\u5927\u5c0f\u662f10\uff0c\u4f9d\u6b21\u4fdd\u5b580-9\u5b57\u7b26\u51fa\u73b0\u4e2a\u6570\uff0c\u904d\u5386\u5b57\u7b26\u4e32\uff0c\u5224\u65ad\u8be5\u5b57\u7b26\u662f\u5426\u662f\u6570\u5b57\u5b57\u7b26\uff0c\u5982\u679c\u662f\u5bf9\u5e94\u7684\u6570\u7ec4\u5143\u7d20\u81ea\u52a01\u3002
\u53c2\u8003\u4ee3\u7801\uff1a
#include#includeint main(){ char a[100];int i,n,num[10];gets(a);n=strlen(a);for(i=0;i='0'&&a[i]<='9')num[a[i]-'0']++;for(i=0;i<10;i++)printf("%d ",num[i]); return 0;}/*\u8f93\u51fa\uff1aads51df0a3df9d87f89ads4fd0f3adsf21sdf47d89f9adsf65as1d2f3asd2 3 2 3 2 2 1 2 3 4*/

#include
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("\u8bf7\u8f93\u5165\u4e00\u4e2a\u5b57\u7b26\u4e32:\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\u7684\u4e2a\u6570\uff1a%d\n",NumCount.num0);
printf("1\u7684\u4e2a\u6570\uff1a%d\n",NumCount.num1);
printf("2\u7684\u4e2a\u6570\uff1a%d\n",NumCount.num2);
printf("3\u7684\u4e2a\u6570\uff1a%d\n",NumCount.num3);
printf("4\u7684\u4e2a\u6570\uff1a%d\n",NumCount.num4);
printf("5\u7684\u4e2a\u6570\uff1a%d\n",NumCount.num5);
printf("6\u7684\u4e2a\u6570\uff1a%d\n",NumCount.num6);
printf("7\u7684\u4e2a\u6570\uff1a%d\n",NumCount.num7);
printf("8\u7684\u4e2a\u6570\uff1a%d\n",NumCount.num8);
printf("9\u7684\u4e2a\u6570\uff1a%d\n",NumCount.num9);

}

思路：统计字符串中每个数字字符出现的个数可以先定义一个整数数组a大小是10，依次保存0-9字符出现个数，遍历字符串，判断该字符是否是数字字符，如果是对应的数组元素自加1。
参考代码：

#include<stdio.h>
#include<string.h>
int main()
{ 
	char a[100];
	int i,n,num[10];
	gets(a);
	n=strlen(a);
	for(i=0;i<10;i++){
		num[i]=0;
	}
	for(i=0;i<n;i++)
		if(a[i]>='0'&&a[i]<='9')
			num[a[i]-'0']++;
	for(i=0;i<10;i++)
	printf("%d ",num[i]);
    return 0;
}
/*
输出：
ads51df0a3df9d87f89ads4fd0f3adsf21sdf47d89f9adsf65as1d2f3asd
2 3 2 3 2 2 1 2 3 4
*/