K-based NumbersTime Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64uSubmit StatusDescriptionLet’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 ≤ 18.InputThe numbers N and K in decimal notation separated by the line break.OutputThe result in decimal notation.Sample Inputinput	output210	90/*************************************************************************
	> File Name: i.cpp
	> Author:yuan 
	> Mail: 
	> Created Time: 2014年11月09日 星期日 21时52分04秒
 ************************************************************************/

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
using namespace std;
long long ans[2][20];
int n,k;
int main()
{
        cin>>n>>k;
        memset(ans,0,sizeof(ans));
        ans[0][1]=0;ans[1][1]=k-1;/*分别指定了长度为1,末尾数字为0的个数为0,不为的个数为k-1*/
        for(int i=2;i<=n;i++)
        {
            ans[0][i]=ans[1][i-1];/*不能出现连续的0,则长度为n末尾为0的个数等于长度为n-1末尾不为0的个数*/
            ans[1][i]=(k-1)*(ans[0][i-1]+ans[1][i-1]);/*长度为n，末尾数字不为0的个数：末尾数字可以是1—k-1中的任意一个
            倒数第二位可以为0,也可以不为0*/
        }
        cout<<ans[0][n]+ans[1][n]<<endl;
    return 0;
}

第一轮 I