PAT (Advanced Level) Practice 1132 Cut Integer (20分) （atoi、stoi区别、stringstream使用）

1.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 <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

2.题目分析

1.atoi、stoi的区别;

atoi是用来将char*转换成int，stoi是将string转换成int

 2.stringstream使用

重复使用要用ss.str(""); ss.clear();清空

3.代码

#include<iostream>
#include<string>
#include<cstring>
#include<sstream>
using namespace std;
int main()
{
	int n;
	long long a, b, c;
	stringstream ss;
	string temp;
	cin >> n;
	for (int i = 0; i<n; i++)
	{
		cin >> temp;
		ss << temp;
		ss >> a;
		ss.str(""); ss.clear();
		ss << temp.substr(0,temp.length()/2);
		ss >> b;
		ss.str(""); ss.clear();
		ss << temp.substr(temp.length() / 2, temp.length() / 2);
		ss >> c;
		ss.str(""); ss.clear();
		if (b*c!=0&&a%( b*c)==0)
			printf("Yes\n");
		else printf("No\n");
	}
}

#include<iostream>
#include<string>
#include<cstring>
#include<sstream>
using namespace std;
int main()
{
	int n;
	long long a, b, c;
	string temp;
	cin >> n;
	for (int i = 0; i<n; i++)
	{
		cin >> temp;
		a = stoi(temp);
		b = stoi(temp.substr(0, temp.length() / 2));
		c = stoi(temp.substr(temp.length() / 2, temp.length() / 2));
		if (b*c!=0&&a%( b*c)==0)
			printf("Yes\n");
		else printf("No\n");
	}
}