大数加法

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此博客提供了一个解决大型整数求和问题的算法，并详细展示了如何通过输入两个大整数，计算它们的和并输出相应的数学方程式。

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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

2
1 2
112233445566778899 998877665544332211

Sample Output

Case 1:
1 + 2 = 3

Case 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;
}