hduoj1002——A + B Problem II

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<stdio.h>#include<string.h>main(){    int k=0;    int f=0;    int i;    int a[1005],b[1005];    char x[1005],y[1005];    int t;    int shu;    int sum[1005];    int n,m;    scanf("%d",&t);    while(t--)    {        k++;        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        memset(sum,0,sizeof(sum));        scanf("%s%s",x,y);        n=strlen(x);        m=strlen(y);        for(i=0;i<n;i++)        {            a[i]=x[n-i-1]-'0';        }        for(i=0;i<m;i++)        {            b[i]=y[m-i-1]-'0';        }        shu=0;        for(i=0;i<1001;i++)        {            sum[i]=a[i]+b[i]+shu;            shu=sum[i]/10;            sum[i]=sum[i]%10;        }        i=1001;        while(sum[i]==0)        {            i--;        }        if(f)        printf("\n");        f=1;        printf("Case %d:\n",k);        printf("%s + %s = ",x,y);        for(;i>=0;i--)        printf("%d",sum[i]);        printf("\n");    }    }