杭电1002 A+Bproblem。。。交了七遍，最后发现原来有五遍是因为没删测试输出= =、、

 

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 98985    Accepted Submission(s): 18771

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
   
   
   
   

 

#include<stdio.h>
#include<string.h>
int main()
{
    int n,i,l1,l2,max,a,j,t;
    char s1[1001],s2[1001],s3[1001];
    scanf("%d",&n);
    for(i = 0;i < n;i++)
    {
        scanf("%s%s",s1,s2);

        l1 = strlen(s1);
        l2 = strlen(s2);
        if(l1>l2)
            max = l1;
        else
            max = l2;

        t=max;
        memset(s3,'\0',1001);
        memset(s3,'0',max+1);
        l1 = l1-1;
        l2 = l2-1;
        max = max;
        s3[0]='0';
        while(l1>=0 && l2>=0)
        {
            a = s1[l1--]-'0'+s2[l2--]-'0'+s3[max]-'0';//问题可以直接简化为小学的算术题，从后往前算
            s3[max--] ='0'+a%10;
            if(a>=10)
                s3[max]+=1;//大于十的时候记得向前加一，因为以上三个数永远不会大于等于20.。。。

        }
        while(l1>=0)
        {
            a = s3[max]-'0'+s1[l1--]-'0';
            s3[max--]='0'+a%10;
            if(a>=10)
                s3[max]+=1;
        }
        while(l2>=0)
        {
            a = s3[max]-'0'+s2[l2--]-'0';
            s3[max--]='0'+a%10;
            if(a>=10)
                s3[max]+=1;
        }

        printf("Case %d:\n",i+1);
        printf("%s + %s = ",s1,s2);

        if(s3[0] > '0')
            printf("%c",s3[0]);//这一步很重要，假设是1和9想加，会上升一位
        for(j = 1;j < t; j++)
            printf( "%c", s3[j]);
        printf( "%c\n", s3[j]);
        if(i<n-1)
            printf("\n");
    }
    return 0;
}