ＨＤＵ　A + B Problem II

2017-11-15 1183

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简介：

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 307 Accepted Submission(s): 147

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

字符串模拟大数加法

#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
char sum[1000];
void add(string a,string b)
{
     string c ;
    if(a.length()<b.length())
    {c = a ; a = b; b = c;}
    int l1 = a.length()-1;
    int l2 = b.length()-1;
    int k=0,i=0;
    memset(sum,0,sizeof(sum));
    char t;
        while(l2!=-1)
        {
           t = (a[l1]-'0')+(b[l2]-'0');
           sum[i]= (t+k)%10+'0';
           k = (t+k)/10;
           l1--;
           l2--;
           i++;
        }
         while(l1!=-1)
         {
             t = (a[l1]-'0');
             sum[i] = (k+t)%10+'0';
             k = (t+k)/10;
             l1--;
             i++;
         }
         if(k!=0)sum[i]=k+'0';
}
int main()
{
    string a,b;
    int T;
    cin>>T;
    for(int i = 1 ; i <= T ;i ++)
    {
        cin>>a>>b;
        add(a,b);
        cout<<"Case "<<i<<":"<<endl;
        cout<<a<<" + "<<b<<" = ";
        for(int j = strlen(sum)-1; j >=0;j--)
        {
            cout<<sum[j];
        }
        if(i<T)
        cout<<endl<<endl;
        else cout<<endl;
    }
    return 0;
}

本文转自NewPanderKing51CTO博客，原文链接： http://www.cnblogs.com/newpanderking/archive/2011/07/31/2122519.html ，如需转载请自行联系原作者

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