GDUFE ACM1002

A+B(Big Number Version)

Time Limit: 2000/1000ms (Java/Others)

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 400.

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:

3
1 2
112233445566778899 998877665544332211
33333333333333333333333333 100000000000000000000

Sample Output:

Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

Case 3:
33333333333333333333333333 + 100000000000000000000 = 33333433333333333333333333
这道题难度中等

核心代码：
       scanf("%s%s",s1,s2);
        len1=strlen(s1);   len2=strlen(s2);                                  
        for(i=len1-1,j=0;i>=0;i--)  //将s1字符串数组转换为数字数组，逆序
            num1[j++]=s1[i]-'0';
        for(i=len2-1,j=0;i>=0;i--)  
            num2[j++]=s2[i]-'0';   
        for(i=0;i<N;i++) 
        {  
            num1[i]+=num2[i];   
            if(num1[i]>9)      //进位
            {   
                num1[i]-=10;       //加法最大是9+9=18
                num1[i+1]++;     //所以加1
            } 
        }           
        printf("Case %d:\n%s + %s = ",X,s1,s2);
        X++;
        for(i=N-1;(i>=0)&&(num1[i]==0);i--);     //空语句去掉多余的0
        for(;i>=0;i--)    
            printf(“%d”,num1[i]);   //逆序输出结果
        

另附一种AC代码：

#include<stdio.h>
#include<string.h>
int main()
{
    char a[1000],b[1000],c[1001];
    int i,j=1,p=0,n,n1,n2;
    scanf("%d",&n);
    while(n)
    {
        scanf("%s %s",a,b);
        printf("Case %d:\n",j);
        printf("%s + %s = ",a,b);
        n1=strlen(a)-1;
        n2=strlen(b)-1;
        for(i=0;n1>=0||n2>=0;i++,n1--,n2--)
        {
            if(n1>=0&&n2>=0){c[i]=a[n1]+b[n2]-'0'+p;}
            if(n1>=0&&n2<0){c[i]=a[n1]+p;}
            if(n1<0&&n2>=0){c[i]=b[n2]+p;}
            p=0;
            if(c[i]>'9'){c[i]=c[i]-10;p=1;}
        }
        if(p==1){printf("%d",p);}
        while(i--)
        {printf("%c",c[i]);}
        j++;
        if(n!=1){printf("\n\n");}
        else {printf("\n");}
        n--;
    }
}