Problem 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
思路很簡單,注意進位即可。
//因為部分細節搞了半天:
//使用max要引用algorithm
//注意最後一個sample不能輸出空行
#include
#include
using namespace std;
int main()
{
char n1[1001] = {0};
char n2[1001] = {0};
int total;
//cin>>total;
scanf("%d", &total);
int outputindex = 0;
while (outputindex < total)
{
//cin>>n1>>n2;
scanf("%s %s", n1, n2);
int len1 = 0;
int len2 = 0;
while ('\0' != n1[len1])
len1++;
while ('\0' != n2[len2])
len2++;
char n[1002] = {0};
int index1 = len1 - 1;
int index2 = len2 - 1;
int index = max(index1, index2) + 1;
char bit = 0;
int num = 0;
while (index1 >= 0 && index2 >= 0)
{
num = n1[index1] + n2[index2] - 48 + bit;
if (num > 57)
{
n[index] = num - 10;
bit = 1;
}
else
{
n[index] = num;
bit = 0;
}
--index1;
--index2;
--index;
}
while (index1 >= 0)
{
num = n1[index1] + bit;
if (num > 57)
{
n[index] = num - 10;
bit = 1;
}
else
{
n[index] = num;
bit = 0;
}
--index1;
--index;
}
while (index2 >= 0)
{
num = n2[index2] + bit;
if (num > 57)
{
n[index] = num - 10;
bit = 1;
}
else
{
n[index] = num;
bit = 0;
}
--index2;
--index;
}
if (bit != 0)
{
n[index] = bit + '0';
}
char *p = n;
while ('\0' == *p)
p++;
++outputindex;
cout<<"Case "<