1132 Cut Integer (20分)--PAT甲级真题

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：给出一个数字Z (10 ≤ Z <2^​31)，保证Z的数字个数为偶数；数字Z可以分成两个整数a和b(比如Z = 167334, 则有 a = 167, b = 334); 如果Z可以被a*b整除，则输出Yes，否则，输出No;

分析：
这里要把整数按照字符顺序分成两部分，最好按照字符串读入Z，然后分割出a和b;

一定一定要注意！！！a/b要保证b不为0，取余运算a%b，也要保证b不为0

#include    
#include
using namespace std; 
int main(){
	int n;
	string arr[22];
	scanf("%d", &n);
	for (int i = 0; i < n; i++)
		cin >> arr[i];
	for (int i = 0; i < n; i++){
		int len = arr[i].length();
		string str1 = arr[i].substr(0, len / 2);
		string str2 = arr[i].substr(len / 2, len / 2); 
		int z = atoi(arr[i].c_str());
		int a = atoi(str1.c_str());
		int b = atoi(str2.c_str()); 
		if ((a*b != 0) && z % (a*b) == 0)
			printf("Yes\n");
		else
			printf("No\n");
	} 
	return 0;
}