只要两个数加法不产生进位就能满足题意。对于首位来说,可以有8*9/2=36种选数方式(因为首位不为0),对于其他位来说,可以有9*10/2=55种选数方式,乘法原理以后,剩下来的就是大数乘法和输出了。
#include<iostream>
using namespace std;
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const int MAX = 100;
class Num
{
public:
Num(int n);
Num operator *(int a);
void Print();
private:
int data[MAX];
};
int main()
{
int n;
while(cin>>n)
{
Num a(36);
for(;n>1;n--) a=a*55;
a.Print();
}
return 0;
}
Num::Num(int n)
{
int t;
data[MAX-1]=n;
for(t=MAX-2;t>=0;t--)
data[t]=0;
}
Num Num::operator *(int a)
{
int t,mem=0;
Num b(0);
for(t=MAX-1;t>=0;t--)
{
b.data[t]=a*data[t]+mem;
mem=b.data[t]/10;
b.data[t]%=10;
}
return b;
}
void Num::Print()
{
int t;
for(t=0;t<MAX-1;t++)
if(data[t]>0) break;
for(;t<MAX;t++)
cout<<data[t];
cout<<endl;
}
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