HDU 1002： A + B Problem II （大数加法）

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

#include<iostream>
#include<algorithm>
#include<string.h>
#include<vector>
#include<cstring>
#include<cstdlib>
using namespace std;
const int maxn = 1024;
char str1[maxn], str2[maxn];
int  array_1[maxn], array_2[maxn];
void sum()
{
    for(int i = strlen(str1) - 1, j = 0; i >= 0; i--)
        array_1[j++] = str1[i] - ‘0‘;
    for(int i = strlen(str2) - 1, j = 0; i >= 0; i--)
        array_2[j++] = str2[i] - ‘0‘;

    for(int i = 0; i < maxn; i++)
    {
        array_2[i] += array_1[i];
        if(array_2[i] >= 10)
        {
            array_2[i + 1] += array_2[i] / 10;
            array_2[i] %= 10;
        }
    }
      cout<<str1<<" + "<<str2<<" = ";
    int k=strlen(str1)>strlen(str2)?strlen(str1):strlen(str2) ;
    for( k=k ; k >= 0 && array_2[k] == 0; k--);
        if(k >= 0)
            for( ; k >= 0; k--)
                cout << array_2[k];
        else
            cout << 0;
    cout << endl;
}

int main()
{
    int t;
    cin>>t;
    for(int i=1;i<=t;i++)
    {

        memset(array_1, 0, sizeof(array_1));
        memset(array_2, 0, sizeof(array_2));
        memset(str1, 0, sizeof(str1));
        memset(str2, 0, sizeof(str2));
        cin >> str1 ;
        cin >> str2 ;
        cout<<"Case "<<i<<":"<<endl;
        sum();
        if(i!=t)
            cout<<endl;
    }

    return 0;
}

原文地址：https://www.cnblogs.com/Shallow-dream/p/11610485.html