Example of C + + fast power and large number modulo algorithm _c language

One, fast power

In fact （a^b）% p , is to ask, (which a,b,p are larger in the range of int) such problems.

First you know the formula for the remainder: （a*b）%p=（a%p*b%p）%p .

So the power is not the accumulation of opportunity, which gives the code:

int fast (int a,int b,int p)

{  long long a1=a,t=1;

  while (b>0)  

  {if (b&1)     /If power B is odd multiply once, because there will be 2 variable even number, (7/2=3)

  t= (t%p) * (a1%p)%p;

  a1= (a1%p) * (a1%p)%p; 

  b/=2;  }

 return (int) (t%p);

}

Two, the large number takes the model

The principle is that the remainder formula:（a+b）%p=(a%p+b%p)%p;

The large number can be seen as each digit multiplied by its own weight and then added to each digit.

such as: 12345678 = (1*10000000) + (2*1000000) +...+8.

The code is as follows:

Char s[200];

#define MOD 10000010;

int main ()

{  while (gets (s))

{  int k=strlen (s), sum=0;

 for (int i=0;i<k;i++)

 sum= (sum*10+s[i]-' 0 ')%mod;  /Of course, if you are worried that sum may overflow, then open the inside again to remove the

 cout<<sum<<endl;

}}

Third, summary

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