URAL 1013 K-based numbers. Version 3

[报告]

   算法同URAL 1009，不过要开高精度——这个精度有点小大……或许开100进制或1000进制才能保证不MLE。

[程序]

// TASK: 1013 K-based numbers. Version 3
#include <cstdlib>
#include <iostream>
#define N 2000
#define len 1000
using namespace std;
typedef int num[len+1];
num f[N+1];
num ls;
inline void addi(int a[],int b[],int s[])
{
       for (long i=len;i>=1;i--) s[i]=a[i]+b[i];
       for (long i=len;i>=1;i--)
       {
           s[i-1]+=s[i]/100;
           s[i]%=100;
       }
}
inline void mult(int a[],long b,int s[])
{
       for (long i=len;i>=1;i--) s[i]=a[i]*b;
       for (long i=len;i>=1;i--)
       {
           s[i-1]+=s[i]/100;
           s[i]%=100;
       }
}
inline void ouot(int ls[])
{
       long i=1;
       while ((i<len)&&(ls[i]<=0)) i++;
       cout << ls[i++];
       for (;i<=len;i++)
           if (ls[i]>9)
              cout << ls[i];
           else cout << "0" << ls[i];
       cout << endl;
}
int main(int argc, char *argv[])
{
    long n,k;
    cin >> n >> k;
    memset(f,0,sizeof(f));
    f[0][len]=1;
    f[1][len]=k-1;
    for (long i=2;i<=n;i++)
    {
        addi(f[i-1],f[i-2],ls);
        mult(ls,k-1,f[i]);
    }
    ouot(f[n]);
 //   system("PAUSE");
    return EXIT_SUCCESS;
}