A + B Problem II—1002

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

分析

  考虑到数据会超出机器整形表示范围，用字符串代替整数来进行加法运算。

C++代码

/*************************************************
Copyright: 武汉大学计算机学院B507
Author: Ryan
Date: 2015-11-25
Description:超大整数加法运算
**************************************************/
#include<iostream>
#include<string>
using namespace std;

int main() {
int n;
string a,b,sum;
//carry表示进位，有进位时为1，没有进位时为0。
int carry=0;
cin>>n;
for(int i=1;i<=n;i++){
cin>>a>>b;
cout<<"Case "<<i<<":"<<endl;
cout<<a<<" + "<<b<<" = ";
sum="";
//对字符串进行扩充是的两个字符串长度一样。
int m = a.size()-b.size();
if(a.size()>b.size()) {
string::iterator it = b.begin();
b.insert(it,m,'0');
}
else if(a.size()<b.size()){
string::iterator it = a.begin();
a.insert(it,-m,'0');
}

for(int j=a.size()-1;j>=0;j--){
int temp = (a[j]-'0')+(b[j]-'0')+carry;
if(temp>=10){
carry=1;
temp=temp-10;
}
else{
carry=0;
}
char c = temp+'0';
sum=c+sum;
}
if(carry){
sum='1'+sum;
}
cout<<sum;

if(i<n)
cout<<endl<<endl;
else
cout<<endl;
}
return 0;
}