1132. Cut Integer (20)

最新推荐文章于 2022-05-18 11:12:51 发布

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最新推荐文章于 2022-05-18 11:12:51 发布

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本文链接：https://blog.youkuaiyun.com/s136424/article/details/78159339

本文介绍了一种算法，用于判断一个指定的整数是否可以通过切割成两个子整数，并且该整数能被这两个子整数的乘积整除。文章提供了具体的输入输出规格，并附带了一个C++实现示例。

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1132. Cut Integer (20)

Cutting an integer means to cut a K digits long 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 x 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

注意点：

1.输入的整数以左边第一个非0的数字为开头，计算整数的长度。

2.防止乘法结果溢出，保险起见使用long long类型定义整数变量。

我的代码：

#include<iostream>
#include<sstream>
using namespace std;
long long digit(long long n)
{
	long long k=0;
	while(n>0)
	{
		n=n/10;
		k++;
	}
	return k;
}
int main()
{
	long long n,i,m1,m2;
	cin>>n;
	while(n--)
	{
		cin>>m1;
		m2=m1;
		long long x=digit(m1),y=0,z=0;
		char a[60];
		stringstream ss;
		ss<<m1;
		ss>>a;
		for(i=0;i<x/2;i++)
			y=y*10+(a[i]-'0');
		for(i=x/2;i<x;i++)
			z=z*10+(a[i]-'0');
		if(y*z!=0)
		{
			if(m2%(y*z)==0) puts("Yes");
			else puts("No");
		}
		else puts("No");
	}
	return 0;
}