Problem Description

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.

 

Code

#include
     
      #include
      
       #include
       
        int main(){
         int i=1,j,t,lena,lenb,len;    int a[1000],b[1000],sum[1001];    char str1[1000],str2[1000];    scanf("%d",&t);    while(i<=t)    {
             memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        memset(sum,0,sizeof(sum));//a、b数组置零        scanf("%s%s",&str1,&str2);        lena=strlen(str1);        lenb=strlen(str2);        for(j=0;j
        
         lenb ? lena:lenb;            for(j=0;j
         
          9)//处理相加大于十的情况             {
                     sum[j]=sum[j]-10;                sum[j+1]++;             }        }        if(sum[len]==0)         {
                 printf("Case %d:\n",i);            printf("%s + %s = ",str1,str2);            for(j=len-1;j>=0;j--)                printf("%d",sum[j]);            if(i!=t)                printf("\n\n");            else                 printf("\n");        }        else         {
                 printf("Case %d:\n",i);            printf("%s + %s = ",str1,str2);            for(j=len;j>=0;j--)                printf("%d",sum[j]);            if(i!=t)                printf("\n\n");            else                 printf("\n");        }        i++;    }    return 0;} Attention
         
        
       
      
     

void *memset(void *s, int ch, n);

函数解释：将s中当前位置后面的n个字节 （typedef unsigned int size_t ）用 ch 替换并返回 s 。

memset：作用是在一段内存块中填充某个给定的值，它是对较大的

或 进行清零操作的一种最快方法。

 

extern unsigned int strlen(char *s);

头文件：string.h

格式：strlen （ 名）

功能：计算给定 的（unsigned int型）长度，不包括'\0'在内

说明：返回s的长度，不包括结束符NUL。