hdu1002 A + B Problem II

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 166634    Accepted Submission(s): 31864

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

 

AC代码

#include<iostream>using namespace std;int main(){int t,i,j,k,m,n,x;char str1[1001],str2[1001];int a[1001],b[1001],c[1001];       scanf("%d",&t);       k=0;while(t--){          if(k!=0)  printf("\n");  k++;  memset(a,0,sizeof(a));  memset(b,0,sizeof(b));          memset(c,0,sizeof(c));  scanf("%s %s",str1,str2);  n=strlen(str1);  m=strlen(str2);  for(j=0,i=n-1;i>=0;i--,j++)  a[j]=str1[i]-'0';          for(j=0,i=m-1;i>=0;i--,j++)  b[j]=str2[i]-'0';  if(n>m)  x=n;  else  x=m;  for(i=0;i<x;i++)  {  c[i]+=a[i]+b[i];   if(c[i]>9)  {  c[i]-=10;  c[i+1]+=1;  }  }   printf("Case %d:\n",k);printf("%s + %s = ",str1,str2);           if(c[i]!=0)  printf("%d",c[i]);  for(j=i-1;j>=0;j--)  printf("%d",c[j]);   printf("\n");}return 0;}

本题为大数A+B，简单题！

谢谢阅读！