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URAL 1009. K-based Numbers(简单递推)

2014-04-24 12:14 295 查看

每个有两种情况：1.上一次是0这次就有k-1种；2.上一次非零这次有k种。

1009. K-based Numbers

Time limit: 1.0 second

Memory limit: 64 MB

Let’s consider K-based numbers, containing exactly N digits. We define a number to be valid if itsK-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 Ndigits.

You may assume that 2 ≤ K ≤ 10; N ≥ 2; N + K ≤ 18.

Input

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

Output

The result in decimal notation.

Sample

input	output
2 10	90

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-8
#define M 1000100
//#define LL __int64
#define LL long long
#define INF 0x3f3f3f3f
#define PI 3.1415926535898

const int maxn = 110;
using namespace std;

int dp[maxn];
int num[maxn];

int main()
{
int n, k;
while(cin >>n>>k)
{
dp[0] = k-1;
for(int i = 1; i < n; i++)
{
num[i] = dp[i-1];
dp[i] = (num[i-1]+dp[i-1])*(k-1);
}
cout<<dp[n-1]+num[n-1]<<endl;
}
}

import java.util.*;
import java.io.*;
import java.math.*;

public class Main
{
public static void main(String args[])
{
Scanner cin = new Scanner(System.in);
BigInteger []dp = new BigInteger[2000];
BigInteger []num = new BigInteger[2000];
while(cin.hasNext()){
int n, k;
n = cin.nextInt();
k = cin.nextInt();
dp[0] = BigInteger.valueOf(k-1);
num[0] = BigInteger.ZERO;
for(int i = 1; i < n; ++i){
num[i] = dp[i-1];
//System.out.println(dp[i-1]);
BigInteger t = num[i-1].add(dp[i-1]);
//System.out.println(t);
dp[i] = t.multiply(BigInteger.valueOf(k-1));
}
System.out.println(dp[n-1].add(num[n-1]));
}
}
}

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