【ACM】nyoj_103_A+BII_201307291022

A+B Problem II
时间限制：3000 ms  |  内存限制：65535 KB
难度：3
描述
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

A,B must be positive.

输入
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.
输出
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.
样例输入
2
1 2
112233445566778899 998877665544332211
样例输出
Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

#include <stdio.h>
#include <string.h>
#define MAX_LEN 1000
int an1[MAX_LEN+100];
int an2[MAX_LEN+100];
char str1[MAX_LEN+100];
char str2[MAX_LEN+100];
int main()
{
 int k,N;
 scanf("%d",&N);
 for(k=1;k<=N;k++)
 {  
  int i,j,len1,len2;
  memset(an1,0,sizeof(an1));
  memset(an2,0,sizeof(an2));
  scanf("%s%s",str1,str2);
  len1=strlen(str1);
  for(j=0,i=len1-1;i>=0;i--)
  an1[j++]=str1[i]-'0';
  len2=strlen(str2);
  for(j=0,i=len2-1;i>=0;i--)
  an2[j++]=str2[i]-'0';
  for(i=0;i<MAX_LEN;i++)
  {
  an1[i]+=an2[i];
  if(an1[i]>=10)
  {
   an1[i]-=10;
   an1[i+1]++;
  }
  }
  printf("Case %d: ",k);
  printf("%s + %s = ",str1,str2);
  for(i=MAX_LEN+100;(i>=0)&&(an1[i]==0);i--);
  if(i>=0)
  for(;i>=0;i--)
  printf("%d",an1[i]);
  else
  printf("0");
  printf(" ");
 }
 return 0;
}

hdu_1002_A+BII_bignum_201307291100

#include <stdio.h>
#include <string.h>
#define MAX_LEN 1000
int an1[MAX_LEN+100];
int an2[MAX_LEN+100];
char str1[MAX_LEN+100];
char str2[MAX_LEN+100];
int main()
{
 int k,N,t=0;
 scanf("%d",&N);
 for(k=1;k<=N;k++)
 {  
  int i,j,len1,len2;
  memset(an1,0,sizeof(an1));
  memset(an2,0,sizeof(an2));
  scanf("%s%s",str1,str2);
  len1=strlen(str1);
  for(j=0,i=len1-1;i>=0;i--)
  an1[j++]=str1[i]-'0';
  len2=strlen(str2);
  for(j=0,i=len2-1;i>=0;i--)
  an2[j++]=str2[i]-'0';
  for(i=0;i<MAX_LEN;i++)
  {
  an1[i]+=an2[i];
  if(an1[i]>=10)
  {
   an1[i]-=10;
   an1[i+1]++;
  }
  }
  printf(t++?" Case %d: ":"Case %d: ",k);
  printf("%s + %s = ",str1,str2);
  for(i=MAX_LEN+100;(i>0)&&(an1[i]==0);i--);  
  for(;i>=0;i--)
  printf("%d",an1[i]);
  printf(" ");
 }
 return 0;
}

较之上个程序有所修改，输出部分有所简化