1002 A + B Problem II

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 205527    Accepted Submission(s): 39513

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

 

#include<iostream>#include<cstring>using namespace std;void add(char *a, char *b){    int i,j,la,lb,max;    int num[1000]={0},aa[1000]={0},bb[1000]={0};    la=strlen(a);   lb=strlen(b);    max=(la>=lb)?la:lb;    strrev(a);      strrev(b);    for(i=0;i<la;i++)        aa[i]=a[i]-48;    for(i=0;i<lb;i++)        bb[i]=b[i]-48;    for(i=0;i<max;i++)    {        j=num[i]+aa[i]+bb[i];        if(j>9) { num[i+1]++; j-=10; }        num[i]=j;    }    if(num[max]) max++;    for(i=max-1;i>=0;i--)        cout<<num[i];    cout<<endl;}int main(){    int i,n;    char a[1000],b[1000];    cin>>n;    for(i=1;i<=n;i++)    {        cin>>a>>b;        if(i!=1) cout<<endl;        cout<<"Case "<<i<<':'<<endl;        cout<<a<<" + "<<b<<" = ";        add(a,b);    }}