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

21 2112233445566778899 998877665544332211

 

Sample Output

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110
一道A+B的题，但是是一道大数题，需用字符串，记得第一次做这道题的时候是在一场训练赛上，当时的水平停留在水题阶段，没有见识过大数题，字符串也学得不怎么样，当时总觉得自己做得没错，但是一直WR，现在再做这道题，感觉蛮轻松，附上代码
#include<stdio.h>#include<string.h>int shu (char a){    return (a-'0');//将字符串转换为数字}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字符串的长度      b1=strlen(b);//b字符串的长度      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不等长的情况下，得出每个a+b直到没有a或者b为止      {          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则把a遗漏的加上      {          num[k--]+=shu(a[--a1]);          if(num[k+1]/10)//注意前面已经k--，求上一个数的时候k+1          {              num[k]++;              num[k+1]=num[k+1]%10;          }      }      while(b1>0)//若b>a则把b遗漏的加上      {          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)//注意若第一个为0不输出          {              i++;          }          printf("%d",num[i]);      }      printf("\n");  }  return 0;}