A + B Problem II

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

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

---

在写这道题的时候发现这题的AC率极低。这题看似简单实际上坑还是挺多的。很多人匆忙下手没有考虑到很多细节部分，导致一直无法AC。
实际上这题应该属于高精度加法的模板题，写法还是有很多种的。我这里提供两种写法。

要注意的点主要有下面几个：
1.不能直接用int类型的变量来直接相加(题目已经说明了结果的长度在1000位以内连long long的范围都超出了)
2.每一组数据处理前要注意初始化变量值为0
3.在处理输入的数字和输出的数字的时候要注意前导0的问题
4.不要遗漏了最后一次的进位
5.最好将char类型的数组转移到int数组中处理会更省力(本文方法二)
6.注意输出格式的问题:
   1)Case # #=1时之前不需要换行 #>=2时之前要加换行符
   2)A + B = SUM 之后要换行

方法一：

#include 
#include 
void Reverse( char *,int );
int main(void)
{
    char a[1003] = {0}; 
    char b[1003] = {0};
    char c[1003] = {0};
    int n = 0;
    int l = 0;
    int la = 0;
    int lb = 0;
    int x = 0;
    int num = 1;
    int zeroid[2] = {0};
    
    scanf("%d",&n);
    for ( int i = 0 ;i < n ;i++ ) {
        zeroid[0] = 0;
        zeroid[1] = 0;
        x = 0;
        memset(a,0,1003);
        memset(b,0,1003);
        memset(c,0,1003);
        scanf("%s %s",a,b);
        while ( a[zeroid[0]] == '0' ) {
            zeroid[0]++;
        }
        while ( b[zeroid[1]] == '0' ) {
            zeroid[1]++;
        }
        la = strlen(a);
        lb = strlen(b);
        l = la > lb ? la:lb;
        Reverse(a,la);
        Reverse(b,lb);
        for ( int j = 0 ;j < l ;j++ ) {
            c[j] = a[j] + b[j] + x;
            x = c[j] / 10;
            c[j] %= 10;
        } 
        c[l] = x;
        if ( num != 1 ) printf("\n");
        printf("Case %d:\n",num);
        for ( int j = la-zeroid[0]-1 ;j >= 0 ;j-- ) {
            printf("%d",(int)a[j]);
        }
        printf(" + ");
        for ( int j = lb-zeroid[1]-1 ;j >= 0 ;j-- ) {
            printf("%d",(int)b[j]);
        }
        printf(" = ");
        zeroid[0] = l; 
        while ( c[zeroid[0]] == 0 ) {
            zeroid[0]--;
        } 
        for ( int j = zeroid[0] ;j >= 0 ;j-- ) {
            printf("%d",(int)c[j]);
        }
        printf("\n");
        num++;
    }
} 
void Reverse( char *a ,int length) {
    char t[1003] = {0};
    for ( int i = length-1 ;i >= 0 ;i-- ) {
        t[length-i-1] = a[i] - 48;
    }
    for ( int i = 0 ;i < length ;i++ ) {
        a[i] = t[i];
    }
} 

方法二：

#include 
#include 
#include 
using namespace std;

int lena,lenb,num,t,i,n,j,lenc;
int a[1020],b[1020],c[1020];
char s1[1020],s2[1020];

int main(){
  scanf("%d",&n);
  for(i=1;i<=n;++i){
   scanf("%s%s",s1,s2); lena=strlen(s1); lenb=strlen(s2);
   memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); num=0;
   memset(c,0,sizeof(c));
   lenc=max(lena,lenb);
   for(j=0;j<=lena-1;++j){a[lena-j-1]=s1[j]-'0';}
   for(j=0;j<=lenb-1;++j){b[lenb-j-1]=s2[j]-'0';}
   for(j=0;j<=lenc;++j){
    c[j]=a[j]+b[j]+num; num=0;
    if(c[j]>=10){
     c[j]-=10; num=1;
    }
   }
   if(c[lenc]==0){--lenc;}
   i==1?printf("Case %d:\n%s + %s = ",i,s1,s2):printf("\nCase %d:\n%s + %s = ",i,s1,s2);
   for(j=lenc;j>=0;--j){printf("%d",c[j]);}
   printf("\n");
  }

  return 0;
}