1132 Cut Integer (20 分)——甲级（atoi函数）

Cutting an integer means to cut a K digits lone integer Z into two integers of (K/2) digits long integers A and B. For example, after cutting Z = 167334, we have A = 167 and B = 334. It is interesting to see that Z can be devided by the product of A and B, as 167334 / (167 × 334) = 3. Given an integer Z, you are supposed to test if it is such an integer.

Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (≤ 20). Then N lines follow, each gives an integer Z (10 ≤ Z <2^​31). It is guaranteed that the number of digits of Z is an even number.

Output Specification:
For each case, print a single line Yes if it is such a number, or No if not.

Sample Input:
3
167334
2333
12345678
Sample Output:
Yes
No
No

题目大意：给定一个数x，其位数wei必定是偶数。将这个数按wei/2对半分，左边为a，右边为b。如果x%(a*b)==0，输出Yes，否则输出No。

注意：有可能会分出为0的情况，而除0或者模0都是没有意义的，所以要特判一下。

代码如下：

/* 常规解法 */
#include <stdio.h>
int main()
{
	int n, x, i;
	scanf("%d", &n);
	while(n--)
	{
		scanf("%d", &x);
		int y = x;
		int wei = 0;
		while(y != 0)
		{
			wei++;
			y /= 10;
		}
		int ten = 1;
		for(i=1; i<=wei/2; i++) ten *= 10;
		int a = x/ten;
		int b = x%ten;
		if(a*b != 0 && x%(a*b) == 0) printf("Yes\n");
		else printf("No\n");
	}
	return 0;
} 

/* 另解:用字符串输入,然后分成两个字串,最后用atoi函数判断 */
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main()
{
	int n;
	char s[15];
	scanf("%d", &n);
	while(n--)
	{
		scanf("%s", s);
		int len = strlen(s), i;
		char s1[15], s2[15];
		for(i=0; i<len/2; i++)
			s1[i] = s[i];
		s1[i] = '\0';
		for(i=len/2; i<len; i++)
			s2[i-len/2] = s[i];
		s2[i-len/2] = '\0';
		int x = atoi(s);
		int a = atoi(s1);
		int b = atoi(s2);
		if(a*b != 0 && x%(a*b) == 0) printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}

总结

1. 浮点错误：可能出现了除0或者模0的情况，要特判。
2. atoi(a)可以将字符串从下标为a开始到’\0’的字串变成数字。（前提是字符串得是数字）