大数（长长以上）和单个整数之间的除法

Question

我正在尝试按照标题中的建议将一个数字除以一个整数，但我很难做到。我正在考虑将大数字的数字插入一个向量，然后从该向量中重复减去所述数字，直到它为空。但是，我找不到执行此操作的代码。这是我到目前为止一直在尝试的：

   #include <fstream>
using namespace std;
int v[100],r[100];  //In v i store the initial number,in r i store the result
int main()
{
FILE*fin=fopen("imp.in","r");
FILE*fout=fopen("imp.out","w");
int n,x,c,w,k,m=2;
fscanf(fin,"%d",&n); //Reads the number of digits the number has
for(;n>0;--n)
{
    fscanf(fin,"%d",&x);
    v[n]=x;
}
fscanf(fin,"%d",&c); //Reads the digit it has to be divided by
while(v[n]!=0)
{
   k=0;
   while(v[1]>=c)
   {
        v[1]-=c;
        ++k;
   }//As long as the first position in vector can be substracted, increase k
   --v[2];
   v[1]+=10;
    /* Because first position will be negative, take one from second                    position and add ten in the first*/
   w=2;
   while(v[w]<0)
   {
       v[w]=9;
       --v[w+1];
       ++w;
   }
   /*When, for example, you substract one from 1000,it will become 999. This loop helps do that*/
   r[1]+=k;
   if(r[1]>9)
   {
       r[1]-=10;
       w=2;
       ++r[2];
       while(r[w]>9)
       {
           ++r[w+1];
           r[w]=0;
           ++w;
           if(w>m)m=w;
       }
   }
   /*If statement and the line of code above it inserts the result into r[100]*/
}
for(;w>0;--w)fprintf(fout,"%d",r[w]);
return 0;
}

Answer 1

您可以使用 GMP 而不是自己从头开始编写。下载 GMP 源代码、构建它以及学习如何使用它比您自己编写它花费的时间更少。

以 2^32 或 2^64 为基数的铅笔和纸除法比 "division by subtraction" 效率高得多，我相信 GMP 使用的算法比这更好。

使用 GMP 你会写：

mpz_class a("134897563047067890568704560457984059182035823590780968094759123590346040967804956029586405960249562895760983475187459073406984560900123895702851034560016405716045613495619456196450196450165901268051620465016405634056104951923845902387581");
std::cout << a / 7 << std::endl;

大数（长长以上）和单个整数之间的除法

Division between a large numbers (Above long long) and a single integer

c++

division

largenumber