hdu 1002:A + B Problem II（大数问题）

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 158615    Accepted Submission(s): 30098

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

 

算法很简单，小学老师就教过，只不过用C++语言实现了一遍。

 

 1 #include <iostream>

 2 

 3 using namespace std;  4 

 5 int main()  6 {  7     char num1[1001],num2[1001],res[1002];  8     int T,i,len1,len2,count,res_digit,digit_sum,C=0;  9     //输入实验次数T

 10     cin>>T;  11     while(T--){     //循环T次

 12         C++;  13         res[0]='0';  14         cin>>num1;  //输入第一个数，存储进num1（char [1001]）

 15         cin>>num2;  //输入第二个数，存储进num2（char [1001]）  16         //计算第一个数长度（位数），存储进len1

 17         for(len1=0;num1[len1]!='\0';len1++);  18         //计算第二个数长度（位数），存储进len2

 19         for(len2=0;num2[len2]!='\0';len2++);  20         //进行相加运算，存储进res（char [1002]）中。  21         //计算结果应该有的位数

 22         if(len1>=len2) res_digit=len1;  23         else res_digit=len2;  24         res[res_digit+1]='\0';  //先给结果附上结束符

 25         len1--;  26         len2--;  27         //循环相加，len1和len2不断--，直到其中一个减到0

 28         count=0;  29         while(len1!=-1 && len2!=-1){  30             if(count==0)  31                 digit_sum=(num1[len1]-'0')+(num2[len2]-'0');  32             else

 33                 digit_sum=(num1[len1]-'0')+(num2[len2]-'0')+count;  34             if(digit_sum>9){    //如果数字之和为2位数

 35                 res[res_digit]=char(digit_sum%10+'0');  36                 count=1;  37  }  38             else{       //如果数字之和为个位数

 39                 res[res_digit]=char(digit_sum+'0');  40                 count=0;  41  }  42             res_digit--;  43             len1--;  44             len2--;  45  }  46 

 47         if(len1==-1 && len2==-1){  48             cout<<"Case "<<C<<':'<<endl;  49             if(count==0){  50                 cout<<num1<<' '<<'+'<<' '<<num2<<' '<<'='<<' ';  51                 //输出结果，不带开头的0

 52                 for(i=1;res[i]!='\0';i++)  53                     cout<<res[i];  54                 cout<<endl;  55  }  56             else{  57                 cout<<num1<<' '<<'+'<<' '<<num2<<' '<<'='<<' ';  58                 //输出结果，带第一个

 59                 res[0]='1';  60                 cout<<res<<endl;  61  }  62  }  63         else if(len1==-1 && len2!=-1){  64             if(count==0){  65                 res[res_digit]=num2[len2];  66                 res_digit--;  67                 len2--;  68  }  69             else{  70                 res[res_digit]=char(num2[len2]-'0'+count+'0');  71                 res_digit--;  72                 len2--;  73  }  74             while(res_digit!=0){  75                 res[res_digit]=num2[len2];  76                 len2--;  77                 res_digit--;  78  }  79             cout<<"Case "<<C<<':'<<endl;  80             cout<<num1<<' '<<'+'<<' '<<num2<<' '<<'='<<' ';  81             for(i=1;res[i]!='\0';i++)  82                 cout<<res[i];  83             cout<<endl;  84  }  85         else{   //len2==-1 && len1!=-1

 86             if(count==0){  87                 res[res_digit]=num1[len1];  88                 res_digit--;  89                 len1--;  90  }  91             else{  92                 res[res_digit]=char(num1[len1]-'0'+count+'0');  93                 res_digit--;  94                 len1--;  95  }  96             while(res_digit!=0){  97                 res[res_digit]=num1[len1];  98                 len1--;  99                 res_digit--; 100  } 101             cout<<"Case "<<C<<':'<<endl; 102             cout<<num1<<' '<<'+'<<' '<<num2<<' '<<'='<<' '; 103             for(i=1;res[i]!='\0';i++) 104                 cout<<res[i]; 105             cout<<endl; 106  } 107         if(T!=0) 108             cout<<endl; 109  } 110     return 0; 111 

112     //最后加一个'\0' 113     //按格式输出res[1002] 114     //如果res[0]有值，则从0输出 115     //如果res[0]是0，则从1输出

116 }

 

Freecode : www.cnblogs.com/yym2013