1002 1002 A + B Problem II大数相加

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 317528    Accepted Submission(s): 61692

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

 

Author

Ignatius.L

原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=1002

大数相加思路非常简单，只要用数组将大数的每一位都存起来，注意进位。。。。。

AC代码：

#include
#include
#define Max 101
void print(char sum[]);
void bigNumAdd(char a[],char b[],char sum[]);
int main()
{
	char a[Max];
	char b[Max];
	char sum[Max];
	gets(a);
	gets(b);
	bigNumAdd(a,b,sum);
	print(sum);
	return 0;
}

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[Max]={0};
	char n[Max]={0};
	memset(sum,0,Max*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=0;j--)
	{
		printf("%c",sum[j]);
	}
}