a+b problem 2

A + B Problem II

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

Limit: 65536/32768 K (Java/Others)Total Submission(s): 3196 Accepted Submission(s): 1178 

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

21 2112233445566778899 998877665544332211

 

Sample Output

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110

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this problem pe

i am very angry!!;

ac daima

#include<iostream>#include<stack>#include<string.h>#include<string>#include <algorithm>using namespace std;int paopao(char x){    if(x=='0')        return 0;    if(x=='1')        return 1;    if(x=='2')        return 2;    if(x=='3')        return 3;    if(x=='4')        return 4;    if(x=='5')        return 5;    if(x=='6')        return 6;    if(x=='7')        return 7;    if(x=='8')        return 8;    if(x=='9')        return 9;}char paopaopao(int x){    if(x==0)        return '0';    if(x==1)        return '1';    if(x==2)        return '2';    if(x==3)        return '3';    if(x==4)        return '4';    if(x==5)        return '5';    if(x==6)        return '6';    if(x==7)        return '7';    if(x==8)        return '8';    if(x==9)        return '9';}int main(){    int t;    cin>>t;    string a;    string b;    int a1,b1;    int pao;    int i,j;    string x;    int pao1;    int pao2;    int pao3;    int k;    k=1;    int qq;    string e1;    string e2;    while(t--)    {        cin>>a>>b;        x="";        string aa;        a1=a.length();        b1=b.length();        e1=a;        e2=b;        reverse(a.begin(),a.end());        reverse(b.begin(),b.end());        if(a1>b1)          {              qq=a1;              a1=b1;              b1=qq;              aa=a;              a=b;              b=aa;          }          pao=0;        for(i=0;i<a1;i++)        {            pao1=paopao(a[i]);            pao2=paopao(b[i]);            pao3=pao1+pao2+pao;            if(pao3>=10)            {                pao=pao3/10;                pao3=pao3%10;            }            else            {                pao=0;            }            x=x+paopaopao(pao3);        }        for(i=a1;i<b1;i++)        {            pao1=paopao(b[i]);            pao3=pao1+pao;            if(pao3>=10)            {                pao=pao3/10;                pao3=pao3%10;            }            else                pao=0;            x=x+paopaopao(pao3);        }        reverse(x.begin(),x.end());        reverse(a.begin(),a.end());        reverse(b.begin(),b.end());        cout<<"Case "<<k++<<":"<<endl;        cout<<e1<<" + "<<e2<<" = ";        if(pao>0)        cout<<paopaopao(pao);        cout<<x;        cout<<endl;       if(t==0)           break;       else            cout<<endl;    }    return 0;}

Author

Ignatius.L