HDU 1002

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 367705    Accepted Submission(s): 71628

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

一道A+B的題，但是是一道大數題，需用字符串，記得第一次做這道題的時候是在一場訓練賽上，當時的水平停留在水題階段，沒有見識過大數題，字符串也學得不怎麼樣，當時總覺得自己做得沒錯，但是一直WR，現在再做這道題，感覺蠻輕鬆，附上代碼

#include<stdio.h>
#include<string.h>
int shu (char a)
{
    return (a-'0');//將字符串轉換爲數字
}

int main()
{
  char a[1000];
  char b[1000];
  int num[10000];
  int s,n,a1,b1,k,i,count=1;
  scanf("%d",&n);
  while(n--)
  {
      if(count!=1)
      {
          printf("\n");
      }
      scanf("%s",&a);
      scanf("%s",&b);
      a1=strlen(a);//a字符串的長度
      b1=strlen(b);//b字符串的長度
      k=(a1>b1)?a1:b1;
      for(i=0;i<=k;i++)
      {
          num[i]=0;//初始化
      }
      s=k;
     for(k;a1>0&&b1>0;k--)//在a，b不等長的情況下，得出每個a+b直到沒有a或者b爲止
      {
          num[k]=num[k]+shu(a[--a1])+shu(b[--b1]);
          if(num[k]/10)
          {
              num[k-1]++;
              num[k]%=10;
          }
      }
      while(a1>0)//若a>b則把a遺漏的加上
      {
          num[k--]+=shu(a[--a1]);
          if(num[k+1]/10)//注意前面已經k--，求上一個數的時候k+1
          {
              num[k]++;
              num[k+1]=num[k+1]%10;
          }
      }
      while(b1>0)//若b>a則把b遺漏的加上
      {
          num[k--]+=shu(b[--b1]);
          if(num[k+1]/10)
          {
              num[k]++;
              num[k+1]=num[k+1]%10;
          }
      }


      printf("Case %d:\n",count++);
       printf("%s + %s = ",a,b);
      for(i=0;i<=s;i++)
      {
          if(i==0&&num[i]==0)//注意若第一個爲0不輸出
          {
              i++;
          }
          printf("%d",num[i]);
      }
      printf("\n");
  }
  return 0;
}