NYOJ103A+B Problem II

A+B Problem II

时间限制：3000 ms  |  内存限制：65535 KB

难度：3

 

描述

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

A,B must be positive.

 

输入

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.

输出

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.

样例输入

2
1 2
112233445566778899 998877665544332211

样例输出

Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

 
/*function:   减法      */
/* date:    2012  0725            */
/*name:    赵杰迪             */
/*environment:Dev-c++ 4.9.9.2 */
#include<stdio.h>
#include<string.h>
#define maxn 1010
int shu1[maxn],shu2[maxn];
char str1[maxn],str2[maxn];
int main()
{
    int n,i,j,len1,len2,k;
    scanf("%d",&n);
    getchar();
    for(k=1;k<=n;k++)
    {
        scanf("%s",str1);
        scanf("%s",str2);
        printf("Case %d:\n%s + %s = ",k,str1,str2);
        memset(shu1,0,sizeof(shu1));
        memset(shu2,0,sizeof(shu2));
        len1=strlen(str1);
        len2=strlen(str2);
        //printf("lenstr1=%d,lenstr2=%d",len1,len2);//
        for(j=0,i=len1-1;i>=0;i--)
            shu1[j++]=str1[i]-'0';
        for(j=0,i=len2-1;i>=0;i--)
            shu2[j++]=str2[i]-'0';
        for(i=0;i<maxn;i++)
        {
            shu1[i]+=shu2[i];
            if(shu1[i]>=10)
            {
                shu1[i]%=10;
                shu1[i+1]++;
            }
        }
    //for(i=0;i<maxn;i++)
        //printf("%d",shu1[i]);//
       for(j=maxn-1;j>=0;j--)
          if(shu1[j])break;
       if(j==-1)
           printf("0");
       else 
        {
            for(i=j;i>=0;i--)
           printf("%d",shu1[i]);
        }
     
     printf("\n");
    }
return 0;
}