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URAL - 1009 - K-based Numbers (简单DP)

时间:2015-03-30 09:30:12      阅读:116      评论:0      收藏:0      [点我收藏+]

标签:acm   ural   dp   

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
Problem Source: USU Championship 1997





AC代码:

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#define LL long long
using namespace std;

int n, k;
LL dp[55];//dp[i]表示位数为i时可以组成的数的个数,注意0在这里可以看做0位数 

int main() {
	while(scanf("%d %d", &n, &k) != EOF) {
		memset(dp, 0, sizeof(dp));
		
		dp[1] = k - 1;
		dp[2] = (k - 1) * k;
		for(int i = 3; i <= n; i++) {
			dp[i] = (k - 1) * (dp[i - 1] + dp[i - 2]);
		}
		
		cout << dp[n] <<endl;
	}
	return 0;
}














URAL - 1009 - K-based Numbers (简单DP)

标签:acm   ural   dp   

原文地址:http://blog.csdn.net/u014355480/article/details/44731071

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