HDU 1002 A + B Problem II

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A + B Problem II

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 12 Accepted Submission(s) : 2

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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

21 2112233445566778899 998877665544332211

Sample Output

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110
大数相加，输入的字符串，最高位在0位置下标处，转化成整形数组的时候要把个位放在0位置下标处，相加的时候不容易出错。下一个坑人的地方是输出。。i==n的时候，只输出回车。 在给定的案例之间才有一行空格。
#include <iostream>#include <stdio.h>#include <memory.h>#include <string>#include <fstream>using namespace std;char s1[1005],s2[1005];int s3[1005],s4[1005],s[1005];int main(void){// freopen("B.txt","r",stdin);   int t;   while(cin>>t)   {       for(int i=1;i<=t;i++)       {           memset(s3,0,sizeof(s3));           memset(s4,0,sizeof(s4));           memset(s,0,sizeof(s));          // printf("Case %d:\n",i);            cin>>s1>>s2;            cout<<"Case "<<i<<":"<<endl;         //  scanf("%s %s",s1,s2);         //  printf("%s + %s = ",s1,s2);          cout<<s1<<" + "<<s2<<" = ";           int l1=strlen(s1);           int k=0;           for(int j=l1-1;j>=0;j--)            s3[k++]=s1[j]-'0';           int l2=strlen(s2);           k=0;           for(int j=l2-1;j>=0;j--)            s4[k++]=s2[j]-'0';           int l=(l1>l2)?l1:l2;           int jin=0;         //  int k=0;           for(int j=0;j<l;j++)           {               s[j]=s3[j]+s4[j]+jin;               jin=s[j]/10;               s[j]=s[j]%10;              // cout<<s[k];             //  k++;           }           int j;           for(j=1005;j>=0;j--)           {               if(s[j])               {                   break;               }           }          // cout<<s;           for(k=j;k>=0;k--)           // printf("%d",s[k]);           cout<<s[k];        //  printf("\n");         cout<<endl;         if(i!=t)          //  printf("\n");            cout<<endl;       }   }    return 0;}

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