九度OJ 1089 數字反轉

題目鏈接：http://ac.jobdu.com/problem.php?pid=1089

題目分析：

使用隊列存放每一位的數字，然後輸出並計算得到反轉數值，從而實現數字反轉。

源代碼：

#include<iostream>
#include<queue>
using namespace std;

int main()
{
	int n;
	cin>>n;
	int m = 1;
	while (m <= n)	//控制行數循環
	{
		int a,b;
		cin>>a>>b;
		int sum = 0,temp = 0;
		queue<int> a1,b1,sum1;
		sum = a + b;
		temp = sum;
		while (sum > 0)
		{
			sum1.push(sum % 10);
			sum = sum / 10;
		}
		while (a > 0)	//將a中每個數字入隊列
		{
			a1.push(a % 10);
			a = a / 10;
		}
		while (b > 0)	//將b中每個數字入隊列
		{
			b1.push(b % 10);
			b = b / 10;
		}
		a = 0, b =0, sum = 0;
		for (int k = 0; k < sum1.size(); k++)
		{
			sum = sum * 10 + sum1.front();
			sum1.pop();
		}
		for (int i = 0; i < a1.size(); i++)
		{
			a = a * 10 + a1.front();
			a1.pop();
		}
		for (int j = 0; j < b1.size(); j++)
		{
			b = b * 10 + b1.front();
			b1.pop();
		}
		if (sum == a + b)
		{
			cout<<temp<<endl;;
		}
		else
		{
			cout<<"NO"<<endl;
		}
		m ++;
	}
	return 0;
}