题目分析：

使用队列存放每一位的数字，然后输出并计算得到反转数值，从而实现数字反转。

源代码：

[cpp]

#include

#include

using namespace std;

int main()

{

int n;

cin>>n;

int m = 1;

while (m <= n)   //控制行数循环

{

int a,b;

cin>>a>>b;

int sum = 0,temp = 0;

queue a1,b1,sum1;

sum = a + b;

temp = sum;

while (sum > 0)

{

sum1.push(sum % 10);

sum = sum / 10;

}

while (a > 0)    //将a中每个数字入队列

{

a1.push(a % 10);

a = a / 10;

}

while (b > 0)    //将b中每个数字入队列

{

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<

}

else

{

cout<

}

m ++;

}

return 0;

}

#include

#include

using namespace std;

int main()

{

int n;

cin>>n;

int m = 1;

while (m <= n) //控制行数循环

{

int a,b;

cin>>a>>b;

int sum = 0,temp = 0;

queue a1,b1,sum1;

sum = a + b;

temp = sum;

while (sum > 0)

{

sum1.push(sum % 10);

sum = sum / 10;

}

while (a > 0) //将a中每个数字入队列

{

a1.push(a % 10);

a = a / 10;

}

while (b > 0) //将b中每个数字入队列

{

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<

}

else

{

cout<

}

m ++;

}

return 0;

}

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