HDU 1002 A + B Problem II (大数加法)

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768
K (Java/Others)

Total Submission(s): 271671 Accepted Submission(s): 52508



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

大数问题，就是用数组模拟计算问题。加法，用数组模拟计算数的加法，进行对每一位进行相加。

附上代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<stdio.h>
#include<string.h>
using namespace std;
int main()
{
char a[1050],b[1050];
int c[1050],d[1050],sum[1050];
int n,num = 0;
cin >> n;
getchar();
while(n--)
{
num++;
memset(c,0,sizeof(c));
memset(d,0,sizeof(d));
memset(sum,0,sizeof(sum));
cin >> a >> b;
int j = 0;
for(int i = strlen(a) - 1;i >= 0;i--)    //存入数组中
{
c[j++] = a[i] - '0';
}
j = 0;
for(int i = strlen(b) - 1;i >= 0;i--)     //存入数组中
{
d[j++] = b[i] - '0';
}
int len = strlen(a) > strlen(b) ? strlen(a) : strlen(b);  //   寻找出最长的数字位数
int jin = 0;
for(int i = 0;i < len;i++)
{
sum[i] = c[i] + d[i] + jin ;
jin = 0;
if(sum[i] > 9 && i < len - 1)         //  i  =  len - 1 的时候为最前面的一位，所以不需要进行进位
{
jin = sum[i]/10;
sum[i] = sum[i]%10;
}
}
if(num != 1)          //  控制输出的格式
cout << endl;
cout << "Case " << num << ":" << endl;
cout << a << " + " << b << " = ";
for(int i = len-1;i >= 0;i--)
{
cout << sum[i];
}
cout << endl;
}
return 0;
}