ACM--steps-2.3.1--大数相加

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 2373 Accepted Submission(s): 918

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

 

Author

Ignatius.L

#include<iostream>
#include<cstring>
using namespace std;
string dyx(string a,string b)
{
    char t;
    string wyx;
    int la=a.length()-1;
    int lb=b.length()-1;
    int flag,i,j;//用以进位。
    i=la;
    j=lb;
    flag=0;
    for(;i>=0&&j>=0;i--,j--)
    {
        t=a[i]+b[j]+flag-'0';
        if(t>'9')
        {
            t=t-10;
            flag=1;
        }
        else
        flag=0;
        wyx+=t;
    }
    //接下来判断若i的位数大于j或j的位数大于i的情况
    while(i>=0)
    {
        t=a[i]+flag;
        if(t>'9')
        {
            t-=10;
            flag=1;
        }
        else
        flag=0;
        wyx+=t;
        i--;
    }
    while(j>=0)
    {
        t=b[j]+flag;
        if(t>'9')
        {
            t-=10;
            flag=1;
        }
        else
        flag=0;
        wyx+=t;
        j--;
    }
    //接下来的判断最后一位如果需要进位的话；
    if(flag)
    wyx+=(flag+'0');
    //换位输出。
    for(i=0,j=wyx.length()-1;i<j;i++,j--)
    {
        t=wyx[i];
        wyx[i]=wyx[j];
        wyx[j]=t;
    }
    return wyx;
}
int main()
{
    int T,n=1;
    cin>>T;
    string a,b;
    while(T--)
    {
        cin>>a>>b;
        cout<<"Case "<<n++<<":"<<endl;
        cout<<a<<" + "<<b<<" = "<<dyx(a,b)<<endl;
        if(T)
        cout<<endl;
    }
    return 0;
}