acm 1002 算法设计

最近突然想往算法方向走走，做了做航电acm的几道题

二话不说，开始

航电acm 1002 题主要是处理长数据的问题，算法原理比较简单，就是用字符数组代替int，因为int太短需要处理的数据较长

下面是问题描述：

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

 

Author

Ignatius.L

下面列出我的代码 参考http://blog.csdn.net/odaynot/article/details/8049632

#include<stdio.h>
#include<string.h>
 int toInt(char c){
     return c-'';
 }
 int main(){
     int i, n,j=,al,bl,ml,t;
     char a[], b[];//存储输入的两个数
    int count[];
     scanf("%d",&n);
     while(n--){
         if(j!=)printf("\n");
         scanf("%s",a);
         al=strlen(a);//返回字符串结束符之前的字符个数
         scanf("%s",b);
         bl=strlen(b);
         ml=(al>bl)?al:bl;//获取较长的字符长度
         t=ml;
         for(i=;i<=ml;i++)count[i]=;//初始化数组
        for(ml;al>&&bl>;ml--){
            count[ml]+=toInt(a[--al])+toInt(b[--bl]);
             if(count[ml]/){
                 //处理进位问题
                 count[ml-]++;
                 count[ml]=count[ml]%;
             }
        }
        while(al>){
            count[ml--]+=toInt(a[--al]);
            if(count[ml+]/){
                count[ml]++;
                count[ml+]%=;
            }
        }
         while(bl>){
             count[ml--]+=toInt(b[--bl]);
             if(count[ml+]/){
                 count[ml]++;
                 count[ml+]%=;
             }
         }

         printf("Case %d:\n%s + %s =",j++,a,b);
         for(i=;i<=t;i++){
             if(i==&&count[i]==){
                 i++;
             }
             printf("%d",count[i]);
         }
         printf("\n");
     }
     return ;
 }

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