如何计算大数的组合 [英] how to calculate combination of large numbers

问题描述

我计算出的数字排列为：-

I calculated permutation of numbers as:-

nPr = n！/（nr）！

其中n和r是给定的。
1 <= n，r< = 100

where n and r are given . 1<= n,r <= 100

i find p=(n-r)+1
and 
for(i=n;i>=p;i--)
  multiply digit by digit and store in array.

但是我将如何计算nCr = n！/ [r！ *（nr）！]的范围相同。？

But how will I calculate the nCr = n!/[r! * (n-r)!] for the same range.?

我使用递归方法如下：-

I did this using recursion as follow :-

#include <stdio.h>
typedef unsigned long long i64;
i64 dp[100][100];
i64 nCr(int n, int r)
{
if(n==r) return dp[n][r] = 1;
if(r==0) return dp[n][r] = 1;
if(r==1) return dp[n][r] = (i64)n;
if(dp[n][r]) return dp[n][r];
return dp[n][r] = nCr(n-1,r) + nCr(n-1,r-1);
}

int main()
{
int n, r;
while(scanf("%d %d",&n,&r)==2)
{
    r = (r<n-r)? r : n-r;
    printf("%llu\n",nCr(n,r));
}
return 0;
}

但范围n <== 100，这对于n不起作用> 60。

but range for n <=100 , and this is not working for n>60 .

推荐答案

考虑使用BigInteger类型的类来表示您的大数字。 BigInteger在Java和C＃（。NET Framework的4+版本）中可用。从您的问题来看，您似乎正在使用C ++（应始终将其添加为标记）。因此，尝试在此处和此处，以获取可用的C ++ BigInteger类。

Consider using a BigInteger type of class to represnet your big numbers. BigInteger is available in Java and C# (version 4+ of the .NET Framework). From your question, it looks like you are using C++ (which you should always add as a tag). So try looking here and here for a usable C++ BigInteger class.

最好的之一我见过建议的用于计算二项式系数的方法是 Mark Dominus 。与其他方法相比，N和K值较大时，溢出的可能性要小得多。

One of the best methods for calculating the binomial coefficient I have seen suggested is by Mark Dominus. It is much less likely to overflow with larger values for N and K than some other methods.

static long GetBinCoeff(long N, long K)
{
   // This function gets the total number of unique combinations based upon N and K.
   // N is the total number of items.
   // K is the size of the group.
   // Total number of unique combinations = N! / ( K! (N - K)! ).
   // This function is less efficient, but is more likely to not overflow when N and K are large.
   // Taken from:  http://blog.plover.com/math/choose.html
   //
   if (K > N) return 0;
   long r = 1;
   long d;
   for (d = 1; d <= K; d++)
   {
      r *= N--;
      r /= d;
   }
   return r;
}

只需将所有长定义替换为BigInt，您就应该做好了。

Just replace all the long definitions with BigInt and you should be good to go.

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