Input: 18887Output:
1 occurs 1 times
7 occurs 1 times
8 occurs 3 times
Input:883822Output:
2 occurs 2 times
3 occurs 1 times
8 occurs 3 times
已解决:
统计数字的位数
给定的数字中字符出现次数统计
疑问:
当数字由输入者自定义位数时,该怎么统计字符出现次数?
提示思路即可,无须代码
Input: 18887Output:
1 occurs 1 times
7 occurs 1 times
8 occurs 3 times
Input:883822Output:
2 occurs 2 times
3 occurs 1 times
8 occurs 3 times
已解决:
统计数字的位数
给定的数字中字符出现次数统计
疑问:
当数字由输入者自定义位数时,该怎么统计字符出现次数?
提示思路即可,无须代码
如果不限制输入的位数,可以当做字符输入,一个 while 读入
#include <stdio.h>
int main(int argc, char *argv[]) {
int nums[10] = {0};
char ch;
while ((ch = getchar()) != '\n') {
if (ch >= 48 && ch <= 57) {
nums[ch - '0']++;
}
}
for (int i = 0; i < 10; ++i) {
printf("%d occurs %d times \n", i, nums[i]);
}
return 0;
}
1 回答821 阅读
564 阅读
itoa + switch
array
提取每一位的数字