acm-A+B Problem II

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A+B Problem II

时间限制：3000 ms | 内存限制：65535 KB

难度：3

描述

I have a very simple problem for you. Given two integers A and B,your job is to calculate the Sum of A + B.

A,B must be positive.

输入

The first line of the input contains an integer T(1<=T<=20)which means the number of test cases. Then T lines follow, eachline consists of two positive integers, A and B. Notice that theintegers are very large, that means you should not process them byusing 32-bit integer. You may assume the length of each integerwill not exceed 1000.

输出

For each test case, you should output two lines. The first line is"Case #:", # means the number of the test case. The second line isthe an equation "A + B = Sum", Sum means the result of A + B. Notethere are some spaces int the equation.

样例输入

21 2112233445566778899 998877665544332211

样例输出

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110

来源

经典题目

我的代码：

#include
#include
#include
using namespace std;
void sum(char str1[],char str2[])
{
int ch[1001]={0};
int len_a=strlen(str1);
int len_b=strlen(str2);
int i=len_a-1;
int j=len_b-1;
int k=0;
int c=0;
while(i>=0||j>=0)
{
  if(i<0)
  {
   ch[k]=str2[j]-'0'+c;
  }
  else if(j<0)
  {
   ch[k]=str1[i]-'0'+c;
  }
  else
  {
   ch[k]=str1[i]-'0'+str2[j]-'0'+c;

  }
  c=ch[k]/10;
  ch[k]=ch[k];
  i--;j--;k++;
}
if(c!=0)
  ch[k]=c;
if(ch[k]!=0) cout<<ch[k];
for(i=k-1;i>=0;i--)
{
  cout<<ch[i];
}

cout<<endl;
}
int main()
{
int t;
//ifstream cin("test.txt");
cin>>t;
int count=0;
char str1[1001],str2[1001];
while(t--)
{
  count++;
  cin>>str1>>str2;
  cout<<"Case"<<count<<":"<<endl;
  cout<<str1<<" +"<<str2<<" = ";
  sum(str1,str2);
}
return 0;
}