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  • OpenJudge 2737 大整数除法

    链接地址:http://bailian.openjudge.cn/practice/2737/

    题目:

    总时间限制:
    1000ms
    内存限制:
    65536kB
    描述
    求2个大的正整数相除的商
    输入
    第1行是测试数据的组数n,每组测试数据占2行,第1行是被除数,第2行是除数。每组测试数据之间有一个空行,每行数据不超过100个字符
    输出
    n行,每组测试数据有一行输出是相应的整数商
    样例输入
    3
    2405337312963373359009260457742057439230496493930355595797660791082739646
    2987192585318701752584429931160870372907079248971095012509790550883793197894
    
    10000000000000000000000000000000000000000
    10000000000
    
    5409656775097850895687056798068970934546546575676768678435435345
    1
    
    样例输出
    0
    1000000000000000000000000000000
    5409656775097850895687056798068970934546546575676768678435435345
    
    提示
    除法比较困难,建议大家最后做这题

    思路:

    套用大整数模板

    代码:

      1 #include <iostream>
      2 #include <cstdio>
      3 #include <string>
      4 using namespace std;
      5 
      6 string sub(string s1,string s2)
      7 {
      8     int flag = 0;
      9     if(s1.length()<s2.length() || (s1.length()==s2.length() && s1 < s2))
     10     {
     11         flag = 1;
     12         string temp=s1;
     13         s1=s2;
     14         s2=temp;
     15     }
     16 
     17 
     18     int i,j;
     19     for(i=s1.length()-1,j=s2.length()-1;i>=0;i--,j--)
     20     {
     21         s1[i]=char(s1[i]-(j>=0?s2[j]-'0':0));   //注意细节
     22         if(s1[i]-'0'<0)
     23         {
     24             s1[i]=char(s1[i] + 10);
     25             s1[i-1]--;
     26         }
     27     }
     28 
     29     for(i = 0; i < s1.length(); i++)
     30     {
     31         if(s1[i] != '0') break;
     32     }
     33     if(i == s1.length()) i = s1.length()-1;
     34     s1 = s1.substr(i);
     35 
     36     if(flag) s1 = "-" + s1;
     37 
     38     return s1;
     39 }
     40 string sum(string s1,string s2)
     41 {
     42     if(s1.length()<s2.length())
     43     {
     44         string temp=s1;
     45         s1=s2;
     46         s2=temp;
     47     }
     48     int i,j;
     49     for(i=s1.length()-1,j=s2.length()-1;i>=0;i--,j--)
     50     {
     51         s1[i]=char(s1[i]+(j>=0?s2[j]-'0':0));   //注意细节
     52         if(s1[i]-'0'>=10)
     53         {
     54             s1[i]=char((s1[i]-'0')%10+'0');
     55             if(i) s1[i-1]++;
     56             else s1='1'+s1;
     57         }
     58     }
     59     return s1;
     60 }
     61 
     62 string div(string s1,string s2)
     63 {
     64     string res = "0";
     65     if(s1.length()<s2.length() || (s1.length()==s2.length() && s1 < s2))
     66     {
     67         return "0";
     68     }
     69 
     70     int i;
     71     int sub_length = s1.length() - s2.length();
     72     for(i = sub_length; i >= 0; i--)
     73     {
     74         string s3(i,'0');
     75         string s4 = s2 + s3;
     76         
     77         do
     78         {
     79             string temp = sub(s1,s4);
     80             if(temp[0] == '-') break;
     81             else
     82             {
     83                 s1 = temp;
     84                 res = sum(res,("1" + s3));
     85             }
     86         }while(1);
     87     }
     88     return res;
     89 }
     90 
     91 int main()
     92 {
     93     int n;
     94     cin>>n;
     95     string str1,str2;
     96     while(n--)
     97     {
     98         cin>>str1>>str2;
     99         cout<<div(str1,str2)<<endl;
    100     }
    101     return 0;
    102 }
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  • 原文地址:https://www.cnblogs.com/mobileliker/p/3516936.html
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