【杭电-oj】-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): 314006    Accepted Submission(s): 60840


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

 

用字符串存比较大的数，然后把数倒着写，因为比较利于计算，此时应注意方法，再判断最高位有没有进位即可。

但是师傅建议我这么写，记下，下道题试试。

初始化：



输出：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
int n,num=0;
char a[1010],b[1010];			//把数当做字符串用a,b存
int l1,l2,l3;
int numa[1010]={0},numb[1010]={0};		 //分别存储a,b准换为数字后的值
scanf("%d",&n);
while(n--)
{
memset (numa,0,sizeof (numa));
memset (numb,0,sizeof (numb));
scanf("%s %s",a,b);
l1=strlen(a);
l2=strlen(b);
l3=max(l1,l2);
for(int i=0;i<l1;i++)
numa[l1-i-1]=a[i]-'0';			//为了方便计算，把数倒着存，注意此处的技巧，会常用
for(int i=0;i<l2;i++)
numb[l2-i-1]=b[i]-'0';
numb[l3]=0;
for(int i=0;i<l3;i++)
{
numb[i]=numa[i]+numb[i];
if(numb[i]>9)
{
numb[i]=numb[i]-10;
numb[i+1]++;
}
}
num++;
printf("Case %d:\n",num);
printf("%s + %s = ",a,b);
if(numb[l3]==0)			//判断l3（最后一位）是否为零
{
for(int i=l3-1;i>=0;i--)	//倒着相加，倒着输出
printf("%d",numb[i]);
printf("\n");
}
else
{
for(int i=l3;i>=0;i--)
printf("%d",numb[i]);
printf("\n");
}
if(n!=0)
printf("\n");
}
return 0;
}