杭电ACM----------1002 A+B problemII

Problem DescriptionI have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B. InputThe 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. OutputFor 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 Input21 2112233445566778899 998877665544332211 Sample OutputCase 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110

本题主要是大数相加问题：

可以用字符数组来实现：

void bigNumAdd(char a[], char b[], char sum[])
{
int i = 0;
int c = 0;//表示进位  
//初始化,对以后位运算有很大帮助！  
char m[MAXSIZE] = { 0 };
char n[MAXSIZE] = { 0 };
memset(sum, 0, MAXSIZE*sizeof(char)); //这里不能写成memset(sum,0,sizeof(sum));原因见注意事项1  
//字符串反转且字符串变数字  
int lenA = strlen(a);
int lenB = strlen(b);
for (i = 0; i < lenA; i++)
{
m[i] = a[lenA - i - 1] - '0';
}
for (i = 0; i < lenB; i++)
{
n[i] = b[lenB - i - 1] - '0';
}
//位运算  
for (i = 0; i < lenA || i < lenB; i++)
{
sum[i] = (m[i] + n[i] + c) % 10 + '0';//得到末位  
c = (m[i] + n[i] + c) / 10;//得到进位  
}
}