hdu 1002 A + B Problem II （大数加法）

A + B Problem II

 

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

 

 

思路：这题除了格式有点坑以外还行，我这里把大数加法用个函数表示，显得更清晰明了......

 

 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<queue>
using namespace std;
#define ll long long 

string Sum(string a,string b)//大数加法 
{
    //补前导零 
    while(a.size()<b.size())
        a.insert(0,"0");
    while(b.size()<a.size())
        b.insert(0,"0");
    
    string ans="";//记录结果 
    
    int jinwei=0,sum,yu;//运算所需 
    
    for(int i=a.size()-1;i>=0;i--)//从末尾算起 
    {
        sum=(a[i]-'0')+(b[i]-'0')+jinwei;
        jinwei=sum/10;
        yu=sum%10;
        ans+=(yu+'0');//加上这一位的余数 
    }
    if(jinwei)//可能多一位 
        ans+=(jinwei+'0');
    
    reverse(ans.begin(),ans.end());//由于是从末尾算起，需要逆置字符串 
    
    return ans;
}

int main()
{
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    
    int T;
    
    cin>>T;
    
    string a,b,s;
    
    for(int k=1;k<=T;k++)
    {
        cin>>a>>b;
        s=Sum(a,b);
        
        cout<<"Case "<<k<<":"<<endl;
        cout<<a<<" + "<<b<<" = "<<s<<endl;
        
        if(k!=T)
            cout<<endl;    
    }
    return 0;
}