A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 372108    Accepted Submission(s): 72510

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<cstdio>#include<cstring>#include<algorithm>using namespace std;char a[10000],b[10000];int aa[10000],bb[10000];int cas;void add(){    int c[10000];    memset(c,0,sizeof(c));    memset(bb,0,sizeof(bb));    memset(aa,0,sizeof(aa));    int i,j,k,t;    int len1,len2;    len1=strlen(a);    len2=strlen(b);    for(i=len1-1,j=0; i>=0; i--,j++)    {        aa[j]=a[i]-'0';    }    for(i=len2-1,j=0; i>=0; i--,j++)    {        bb[j]=b[i]-'0';    }    int maxline;    maxline=max(len1,len2);    for(i=0; i<maxline; i++)    {        c[i]=aa[i]+bb[i];    }    for(i=0; i<maxline; i++)    {        if(c[i]>=10)        {            c[i+1]++;            c[i]=c[i]%10;        }    }    for(j=10000-1; j>=0; j--)    {        if(c[j]!=0)        {            break;        }    }    printf("%s",a);    printf(" + ");    printf("%s",b);    printf(" = ");    for(i=j; i>=0; i--)    {        printf("%d",c[i]);    }    printf("\n");}int main(){    int i,j,k,t;    scanf("%d",&t);    cas=1;    while(t--)    {        scanf("%s %s",a,b);        printf("Case %d:\n",cas++);        add();        if(t!=0)        {            printf("\n");        }    }}