大数加法

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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>using namespace std;int c[1100];void zhuan(char *a,int *x){    int len,i;    len=strlen(a);    for(i=0;i<len;i++){        x[len-i]=a[i]-'0';     }}void jia(int *x,int *y){    for(int i=0;i<1001;i++){        c[i]=x[i]+y[i];        if(c[i]>9){            c[i]-=10;            x[i+1]+=1;        }    }}void ptr(int *c){    int i=1001;    for(;i>1;i--){      if(c[i])          for(;i>1;i--){            cout << c[i];        }    }      cout << c[1] << endl;}int main(){    int x[1001],y[1001];    char a[1001],b[1001];    int t,i=0;    cin >> t;    while(t--){        i++;        memset(x,0,sizeof(x));        memset(y,0,sizeof(y));        memset(c,0,sizeof(c));        cin >> a >> b;        printf("Case %d:\n",i);        printf("%s + %s = ",a,b);        zhuan(a,x);        zhuan(b,y);        jia(x,y);        ptr(c);        if(t)        cout << endl;    }    return 0;}