I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

 

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.

 

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.

 

2

1 2

112233445566778899 998877665544332211

 

Case 1:

1 + 2 = 3

Case 2:

112233445566778899 + 998877665544332211 = 1111111111111111110

 

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高精度加法计算，数值位数太大，用字符处理，按位计算，为了方便末尾对齐，采取先取反再相加在取反；

 

#include<iostream>

#include<cstdio>

#include<cstring>

#include<string>

#include<cmath>

using namespace std;

void un(char *a,int k)   //取反

{

    for(int i=0;i<k/2;i++)

        swap(a[i],a[k-1-i]);

}

int main()

{

    char a[1005],b[1005],c[1005],A[1005],B[1005];

    int k1,k2,k;

    int o;

    while(~scanf(" %d",&o))

    {

        int t=0;

    int q=o;

        while(o--)

        {

            memset(c,'0',sizeof(c));

            scanf(" %s %s",A,B);

            strcpy(a,A);

            strcpy(b,B);

            k1=strlen(a);

            k2=strlen(b);

            un(a,k1);

            un(b,k2);

            if(k1>k2)//补零以对齐；

            {

                k=k1;

                for(int i=k2;i<k;i++)

                    b[i]='0';

            }

            else

            {

                k=k2;

                for(int i=k1;i<k;i++)

                    a[i]='0';

            }

            for(int i=0;i<k;i++)//相加和进位

            {

                c[i]=c[i]+(a[i]-'0')+(b[i]-'0');

                if(c[i]>'9')

                {

                    c[i]-=10;

                    c[i+1]+=1;

                }

            }

            printf("Case %d:\n%s + %s = ",++t,A,B);

            if(c[k]>'0')   //最高位进位单独考虑

                printf("%c",c[k]);

            for(int i=k-1;i>=0;i--)

            {

                printf("%c",c[i]);

            }

            printf("\n");

            if(t!=q)

                printf("\n");

        }

    }

    return 0;

}