题目如下：
Description
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.
Output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.
Sample Input
2
1 2
112233445566778899 998877665544332211
Sample Output
Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
在vs上运行没有问题，但是oj上一直结果为Runtime Error (ACCESS_VIOLATION)， _请问为什么？
#include<iostream>
using namespace std;
char a[1010] = { 0 };
char b[1010] = { 0 };
bool flag = true;
int main()
{
int T;
while(cin >> T)
{
if (T < 1 || T > 20)
break;
for (int j = 1; j <= T; j++)
{
if (T == 0)
break;
//below
cin >> a;
cin >> b;
int m = 0, n = 0;//count
for (int i = 0; i < 1000; i++)
{
if (a[i] != 0)
m++;
else
break;
}
for (int i = 0; i < 1000; i++)
{
if (b[i] != 0)
n++;
else
break;
}
char*c = new char[m + 1];//c[m]
char*d = new char[n + 1];//d[n]
int t = 0;
for (int i = m - 1; i >= 0; i--)
{
c[t] = a[i];
t++;
}
c[m] = 0;
t = 0;
for (int i = n - 1; i >= 0; i--)
{
d[t] = b[i];
t++;
}
d[n] = 0;//two integers ready
char*c1 = new char[m + 1];
char*d1 = new char[n + 1];
t = 0;
for (int i = m - 1; i >= 0; i--)
{
c1[t] = c[i];
t++;
}
t = 0;
for (int i = n - 1; i >= 0; i--)
{
d1[t] = d[i];
t++;
}
c1[m] = 0;
d1[n] = 0;
cout << "Case " << j << ":" << endl;
cout << c1 << " + " << d1 << " = ";
int k;
if (m > n)
k = m;
else
k = n;
char*e = new char[k + 1];//extra number
char*f = new char[k + 1];//final number
for (int i = 0; i < k; i++)
{
char z = 0;
if (c[i] > 0 && d[i] > 0)
z = (c[i] - '0') + (d[i] - '0');
else if (c[i] <= 0)
z = d[i] - '0';
else
z = c[i] - '0';
f[i] = z;
if (z > 9)
{
e[i + 1] = '1';
f[i] = z - 10;
}
if (e[i] == '1')
f[i]++;
if (f[i] > 9)
{
e[i + 1] = '1';
f[i] = f[i] - 10;
}
f[i] += '0';
}
if (e[k] == '1')
{
f[k] = '1';
for (int i = k; i >= 0; i--)
{
while (f[i] == '0')
{
i--;
}
if (i < 0)
i++;
cout << f[i];
}
}
else
{
for (int i = k - 1; i >= 0; i--)
{
while (f[i] == '0')
{
i--;
}
if (i < 0)
i++;
cout << f[i];
}
}
cout << endl;
if (j != T)
cout << endl;
delete[]c;
delete[]d;
delete[]c1;
delete[]d1;
delete[]e;
delete[]f;
//above
}
}
return 0;
}
尝试将
char*c = new char[m + 1];//c[m]
char*d = new char[n + 1];//d[n]
改成
char*c = new char[1010];
char*d = new char[1010];