杭电1002

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 364211    Accepted Submission(s): 70857

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

就是大数的加减
//此题的要点在于 把数字当字符串储存 用数组储存
//计算的时候注意要减去‘0’，因为那是我们的字符串
//在我们换算 的时候一定要加上temp 我就是因为没加 错了一个下午
#include <iostream>
#include<string>
using namespace std;

int main()
{
     int n;
     cin>>n;
      int ui=1;
     while(n--){
     string a,b;
     int sum[1001];//用来放结果
    cin>>a>>b;

     int lena,lenb;

     lena=a.length();
     lenb=b.length();
    // cout<<lena<<" "<<lenb<<endl;

     int ka=lena-1;
     int kb=lenb-1;
     int q=0;//用来表示数组下标
     int temp=0;//用来表示进位
    // cout<<a[2]+a[1]-'0'-'0'<<endl;

    while(lena!=0&&lenb!=0)
     {
         int numa=a[ka]-'0';//和 因为string是字符类 要减去0的ascii码
         int numb=b[kb]-'0';

       //cout<<numa<<" " <<numb<<endl;
    sum[q]=(numa+numb+temp)%10;//我们要的是余数位
        if((numa+numb+temp)>=10){  //这里的每一次判断必须加上temp进位
            temp=1;
        }else{
            temp=0;
        }
        q++;
        ka--;
        kb--;
        lena--;
        lenb--;
     }

     //cout<<a[ka]<<endl;

     if(lena>lenb){
        while(lena){
        int numa=a[ka]-'0' ;
        sum[q]=(numa+temp)%10;//我们要的是余数位

       temp=(numa+temp)/10;
         q++;
        ka--;
       // kb--;
        lena--;
        //lenb--;
        }
     }

     else if(lenb>lena){
        while(lenb){
            int numb=b[kb]-'0';
            sum[q]=(numb+temp)%10;

           temp=(numb+temp)/10;
            q++;
            kb--;
            lenb--;
        }
     }

   else {
   if(temp==1){
            sum[q]=1;
            q++;
        }

   }


    cout<<"Case "<<ui<<":"<<endl;
    ui++;
    int alll=a.length();
    int blll=b.length();
    for(int j=0;j<alll;j++)
    {
        cout<<a[j];
    }
    cout<<" + ";
    for(int jk=0;jk<blll;jk++)
    {
        cout<<b[jk];
    }
    cout<<" = ";
    for(int i=q-1;i>=0;i--){
        cout<<sum[i];

    }

    if(n==0){
        cout<<endl;
    }
    else{
    cout<<endl;
    cout<<endl;
    }

}

    return 0;
}

然而，我花了一个下午。其实就是把temp给弄错了

错误代码为：

while(lena!=0&&lenb!=0)
     {
         int numa=a[ka]-'0';//和 因为string是字符类 要减去0的ascii码
         int numb=b[kb]-'0';

       //cout<<numa<<" " <<numb<<endl;
    sum[q]=(numa+numb+temp)%10;//我们要的是余数位
        if((numa+numb)>=10){  //就是这边错了！！！！！！
            temp=1;
        }else{
            temp=0;
        }
        q++;
        ka--;
        kb--;
        lena--;
        lenb--;
     }

bug为：