K-based Numbers. Version 2

Description

Let’s consider K-based numbers, containing exactly N digits. We define a number to be valid if its K-based notation doesn’t contain two successive zeros. For example:

• 1010230 is a valid 7-digit number;
• 1000198 is not a valid number;
• 0001235 is not a 7-digit number, it is a 4-digit number.

Given two numbers N and K, you are to calculate an amount of valid K based numbers, containing N digits.

You may assume that 2 ≤ K ≤ 10; N ≥ 2; N + K ≤ 1800.

Input

The numbers N and K in decimal notation separated by the line break.

Output

The result in decimal notation.

Sample Input

input	output
2 10	90

因为N+K<=1800，所以要用到高精度加法还有乘法，至于规律还是跟前一道一样。因为加法写错WA了好多次= =

#include<iostream>
#include<string>
#include<cstring>

using namespace std;

int digit[10000];

string stri[10000];

string mul(string str,int a){
    memset(digit,0,sizeof(digit));
	int jinwei=0,n=0;
    int len=str.length();
    for(int i=len-1; i >= 0; i--){
        digit[n++]=((str[i]-'0')*a+jinwei)%10;
        jinwei=((str[i]-'0')*a+jinwei)/10;
    }
    while(jinwei!=0){
        digit[n++]=jinwei%10;
        jinwei /= 10;
    }
    string st;
    for(int i=n-1; i >= 0; i--)
    	st+=(digit[i]+'0');
    return st;
}
string add(string st1,string st2){
	int len1=st1.length()-1;
	int len2=st2.length()-1;
	int jinwei=0,n=0;
	memset(digit,0,sizeof(digit));
	while(len1>=0&&len2>=0){
		digit[n++]=(st1[len1]-'0'+st2[len2]-'0'+jinwei)%10;
		jinwei=(st1[len1]-'0'+st2[len2]-'0'+jinwei)/10;
		len1--;
		len2--;
	}
	for(int i=len1; i >= 0; i--){
		digit[n++]=(st1[i]-'0'+jinwei)%10;
		jinwei=(st1[i]-'0'+jinwei)/10;
	}
	for(int i=len2; i >= 0; i--){
		digit[n++]=(st2[i]-'0'+jinwei)%10;
		jinwei=(st2[i]-'0'+jinwei)/10;
	}
	string st;
	if(jinwei!=0) st+=(jinwei+'0');
    for(int i=n-1; i >= 0; i--)
    	st+=(digit[i]+'0');
    return st;

	
}
string tostr(int i){
	int n=0;
	while(i!=0){
		digit[n++]=i%10;
		 i/=10;
	}
	string st;
    for(int i=n-1; i >= 0; i--)
    	st+=(digit[i]+'0');
    return st;
}

int main()
{
	
	int n;
	int k;
	cin >> n >> k;
	stri[1]=tostr(k-1);
	stri[2]=mul(tostr(k),k-1);
	for(int i=3; i <= n; i++)
		stri[i]=mul(add(stri[i-1],stri[i-2]),k-1);
	cout << stri[n] << endl;
}