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

Sample Output

Author

Ignatius.L

 1 #include <iostream>
 2  using namespace std;
 3  void add(char a[],char b[])
 4  {
 5    char sum[1010]={' '};
 6    int flg=0;
 7    int temp =0;
 8    int len_a =strlen(a);
 9    int len_b =strlen(b);
10    int i=len_a;
11    int j=len_b;
12    for (;i>0;i--)
13    {
14      if (j>0)
15      {
16        temp =a[i-1]+b[j-1]+flg-96;　　　　//写成temp=a[i-1]-'0'+b[i-1]-'0'+flg  会更好理解，这个语句是用来将temp对应的ascii码真正变为数字，从而用来确定是否进位。
17        j--;
18      }
19      else temp = a[i-1]+flg-48;
20      if (temp>=10)
21      {
22       flg=1;
23      }
24      else flg =0;
25      temp =temp%10;
26      sum[i]=temp+48;
27    }
28   if (flg==1)sum[0]=49;　　　　　　　　　　//到这里前面的循环结束了，所以如果这时候flg为1的话，说明要进位，也就是最高位要再进一位。
29   i=0;
30   while (i<=len_a)
31   {
32     if (sum[i]!=' ')cout<<sum[i];　　　　//在前面预留了一位，用来存放进位，如果没有进位的话，就不会输出。巧妙
33     i++;
34   }
35     cout<<endl;
36   }
37  void main()
38  {
39    int N;
40    while (cin >>N )
41    {
42      for (int i=1;i<=N;i++)
43      {
44        char a[1000];
45        char b[1000];
46        cin >>a;
47        cin >>b;
48        int len_a =strlen(a);
49        int len_b =strlen(b);
50        cout <<"Case "<<i<<":\n"<<a<<" + "<<b<<" = ";
51        if (len_a>=len_b)
52        {
53          add(a,b);
54        }
55        else add(b,a);
56        if (i!=N)cout<<endl;
57       }
58     }
59  }

 1 #include<stdio.h>
 2 #include<string.h>
 3 void add(char a[], char b[])
 4 {
 5     char c[1002]={' '};
 6     int len_a=strlen(a), len_b=strlen(b);
 7     int i=len_a;
 8     int j=len_b;
 9     int temp=0, flag=0;
10
11         for( ; i>0 ; i--)
12         {
13             if(j>0)
14             {
15                 temp=a[i-1]-'0'+b[j -1]+flag-'0';
16                 j--;
17             }
18             else
19                 temp=a[i-1]+flag-'0';
20             if(temp>=10)
21                 flag=1;
22             else
23                 flag=0;
24             temp=temp%10;
25             c[i]=temp+'0';
26         }
27
28     if(flag==1) c[0]='1';
29     for(i=0; i<=len_a; i++)
30     {
31         if(c[i]!=' ') printf("%c",c[i]);
32     }
33     printf("\n");
34 }
35
36 int main()
37 {
38     char a[1001], b[1001];
39     int i, j, n;
40     int len_a, len_b;
41     scanf("%d", &n);
42     for(i=0;i<n;i++)
43     {
44         scanf("%s", &a);
45         len_a=strlen(a);
46         scanf("%s", &b);
47         len_b=strlen(b);
48         printf("Case %d:\n%s + %s = ", i+1, a, b);
49         if(len_a>=len_b)
50             add(a,b);
51         else
52             add(b,a);
53         if(i+1!=n)
54             printf("\n");
55
56     }
57     return 0;
58 }

 1 #include <string>
 2 #include <iostream>
 3 using namespace std;
 4 void main ()
 5 {
 6     int n,i,j=1;
 7     char a[5000],b[5000],c[5000];
 8     scanf ("%d",&n);
 9     getchar ();
10     while (n--)
11     {
12
13         for (i=0;i<5000;i++)
14             c[i]='0';
15         scanf ("%s%s",a,b);
16         strrev (a);
17         strrev (b);                　　　　　　　　　　//倒转数组，使之对齐
18         for (i=0;a[i]!='\0'&&b[i]!='\0';i++)        //进行计算
19         {    if ((a[i]+b[i]+c[i]-96-48)<10)
20                 c[i]=a[i]+b[i]+c[i]-48-48;
21             else
22             {
23                 c[i]=a[i]+b[i]+c[i]-48-58;　　　　　　//减掉一个10
24                 c[i+1]=49;　　　　　　　　　　　　　　　//加上一个1
25             }
26         }
27         if (strlen(a)>strlen(b))        　　　　　　　//看看两个数组哪个长度长，长的那个便是没有相加到最后一位的那个
28             for (;a[i]!='\0';i++)
29                 c[i]=a[i]+c[i]-48;
30         else if (strlen(a)<strlen(b))
31             for (;b[i]!='\0';i++)
32                 c[i]=b[i]+c[i]-48;
33         if (c[i]=='0')
34             c[i]='\0';
35         else c[i+1]='\0';
36         strrev (c);
37         strrev (a);
38         strrev (b);
39         if (n!=0)
40         cout<<"Case "<<j<<":"<<endl<<a<<" + "<<b<<" = "<<c<<endl<<endl;
41         else {cout<<"Case "<<j<<":"<<endl<<a<<" + "<<b<<" = "<<c<<endl;}
42         j++;　　　　　　　　　　//控制 Case N；
43     }
44 }
45 　　　　　　　　　　　　　　　　　　　　　　　　// strrev()  把字符串s的所有字符的顺序颠倒过来
46  

#include<iostream>
#include<string.h>
using namespace std;#define MAX 1010
int add1[MAX], add2[MAX], res[MAX];
char tmp1[MAX], tmp2[MAX];
int main()
{int N, i, j, len, len1, len2, tmp, k;scanf("%d",&N);getchar();for(j=0;j<N;j++){memset(add1,0,sizeof(add1));memset(add2,0,sizeof(add2));memset(res,0,sizeof(res));memset(tmp1,0,sizeof(tmp1));memset(tmp2,0,sizeof(tmp2));scanf("%s %s",tmp1,tmp2);len1 = strlen(tmp1);len2 = strlen(tmp2);for(i=len1-1,k=0;i>=0;i--)add1[k++] = tmp1[i] - '0';for(i=len2-1,k=0;i>=0;i--)add2[k++] = tmp2[i] - '0';tmp = 0;if(len1 >= len2){for(i=0;i<=len1;i++){res[i] = (add1[i] + add2[i] +tmp)%10;tmp = (add1[i] + add2[i] +tmp)/10;}}else if(len1 < len2){for(i=0;i<=len2;i++){res[i] = (add1[i] + add2[i] +tmp)%10;tmp = (add1[i] + add2[i] +tmp)/10;}}if(len1 >= len2) len = len1;else len = len2;printf("Case %d:\n%s + %s = ",j+1, tmp1 , tmp2);if(res[len]!=0) printf("%d",res[len]);    for(i=len-1;i>=0;i--)printf("%d",res[i]);printf("\n");if(j!=N-1) printf("\n");}return 0;
}