HDOJ-1002

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

2 1 2 112233445566778899 998877665544332211

Sample Output

Case 1: 1 + 2 = 3

Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110

 

 大数相加的问题，开两个数组分别储存a、b，再开一个数组sum用来储存最终的和值。

这题要注意相加时储存的顺序和最后输出时的顺序不要搞错。

当然，这里特别强调一下最后的输出格式，最后一组数据只要换一行！一行！行！原谅愚蠢如我在这个点被坑了5次！！还是刷题太少。。

附AC代码

#include<cstdio>
#include<cstring>
const int MAX=1050;
int main()
{
    int n,m,len1,len2,maxlen,minlen;
    char a[MAX],b[MAX],sum[MAX];//三个主要数组 
    char *max, *min;
    scanf("%d",&n);
    for(m=1;m<=n;m++)
    {
        int i,j,flag=0;
        memset(sum,0,sizeof(sum));
        scanf("%s %s",&a,&b);
        len1=strlen(a);
        len2=strlen(b);
        if(len1>=len2)//判断长度，长的为max 
        {
            maxlen=len1;
            minlen=len2;
            max = a;
            min = b;
        }
        else
        {
            maxlen=len2;
            minlen=len1;
            max = b;
            min = a;
        }
        for(i=maxlen-1,j=minlen-1;i>=maxlen-minlen&&j>=0;i--,j--)//长度重叠部分相加 
        {
            sum[i]+=max[i]+min[j]-'0'-'0';//转换为十进制整数，下面的同理 
            if(i==0&&sum[0]>9)//如果两数组长度相等 
            {
                flag=1;//加在sum前表示进的1 
                sum[i]-=10;//本大于9，进1故减10 
            }
            else if(sum[i]>9)
            {
                sum[i-1]++;//进1 
                sum[i]-=10;
            }
        }
        for(i=maxlen-minlen-1;i>=0;i--)//剩下的max加到sum里 
        {
            sum[i]+=max[i]-'0';
            if(i==0&&sum[0]>9)
            {
                flag=1;
                sum[i]-=10;
            }
            else if(sum[i]>9)
            {
                sum[i-1]++;
                sum[i]-=10;
            }
        }printf("Case %d:
",m);
        
        if(flag)        
        {
            
            printf("%s + %s = 1",a,b);}
        else
        {
            
            printf("%s + %s = ",a,b);
        }
        for(i=0;i<maxlen;i++)           
            printf("%d",sum[i]);
        if(m!=n){            //注意这里最后一组的输出 
            printf("

");
        }
        else{
            {printf("
");}
        } 
    }
    return 0;
}