【笨方法学PAT】1132 Cut Integer （20 分）

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探讨了如何通过编程判断一个整数是否可以被其切割后的两个子整数的乘积整除的问题，使用C++实现，涉及字符串操作和条件判断。

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一、题目

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 <231). 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:

Sample Output:

Yes
No
No

二、题目大意

一个整数从中间分成两个整数，原来的整数是否能整出两个新的整数。

三、考点

string

四、注意

1、b*c==0判断，否则浮点错误；

2、substr的用法，substr(n1,n2)，从n1开始的n2个字符。

五、代码

#include<iostream>
#include<string>
using namespace std;
int main() {
	//read
	int n;
	cin >> n;

	//solve
	while (n--) {
		long long int a;
		cin >> a;
		string s =to_string(a);
		int b = stoi(s.substr(0, s.length() / 2));
		int c = stoi(s.substr(s.length() / 2));
		if (b*c != 0 && a % (b*c) == 0)
			cout << "Yes" << endl;
		else
			cout << "No" << endl;
	}

	system("pause");
	return 0;
}