hdu1002——A + B Problem II

原题：

[align=left]Problem Description[/align]
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

分析：

模拟进制，字符串处理；

原码：

#include<stdio.h>
#include<string.h>
int main()
{
char a1[1002],a2[1002];
int c,i,j,n,m,s1[1002],s2[1002],l1,l2,count;
count=1;
scanf("%d",&n);
m=n;
while(m--)
{
memset(s1,0,sizeof(s1));
memset(s2,0,sizeof(s2));//s1、s2全部赋值为0；
scanf("%s%s",a1,a2);
l1=strlen(a1);
l2=strlen(a2);
c=0;
for(i=l1-1; i>=0; i--)
{
s1[c++]=a1[i]-'0';
}
c=0;//将字符串转换成数字
for(i=l2-1; i>=0; i--)
{

s2[c++]=a2[i]-'0';
}//将字符串转换成数字
for(i=0; i<1002; i++)
{
s1[i]+=s2[i];//逐位相加
if(s1[i]>=10)//判断是否进位
{
s1[i]-=10;
s1[i+1]++;//进位
}
}
printf("Case %d:\n",count++);
printf("%s + %s = ",a1,a2);
for(i=1001; i>=0; i--)
{
if(s1[i])
break;
}   //跳出多余的0；
for(j=i; j>=0; j--)
{

printf("%d",s1[j]);//已经跳出多余的0，依次输出。
}
printf("\n");
if(count!=n+1)
printf("\n");
}
return 0;
}