ural 1009. K-based Numbers - dp
2016-11-24 23:57
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1009. K-based Numbers
Let’s consider K-based numbers, containing exactlyN 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 validK based numbers, containing
N digits.
You may assume that 2 ≤ K ≤ 10; N ≥ 2;N +
K ≤ 18.
Sample
Problem Source: USU Championship 1997
题目大意
长度为n的k进制数,问没有两个相邻0的个数有几个
解题思路
因为长度为n,所以最高位肯定是不为零的,并且我们能猜到,长度为n的个数和长度为n-1的个数有关。
一开始想的是前面的位数都确定下来了,最后一位放在最后解决,但是不能确保倒数第二位为不为零。
但是最高位肯定不为0了,那么我们每次把新的位插在前面,这样状态转移方程
dp[i] = dp[i-1]*(k-1) + dp[i-2]*(k-1)
dp[i-1]*(k-1)是因为最高位不能为0,而长度为i-1的时候已经保证了第二高位不为0,所以只有k-1种选择
dp[i-2]*(k-1) 是因为 最高位不为0有k-1种选择,第二高位可以为0,有一种选择
代码
Let’s consider K-based numbers, containing exactlyN 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 validK based numbers, containing
N digits.
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 |
题目大意
长度为n的k进制数,问没有两个相邻0的个数有几个
解题思路
因为长度为n,所以最高位肯定是不为零的,并且我们能猜到,长度为n的个数和长度为n-1的个数有关。
一开始想的是前面的位数都确定下来了,最后一位放在最后解决,但是不能确保倒数第二位为不为零。
但是最高位肯定不为0了,那么我们每次把新的位插在前面,这样状态转移方程
dp[i] = dp[i-1]*(k-1) + dp[i-2]*(k-1)
dp[i-1]*(k-1)是因为最高位不能为0,而长度为i-1的时候已经保证了第二高位不为0,所以只有k-1种选择
dp[i-2]*(k-1) 是因为 最高位不为0有k-1种选择,第二高位可以为0,有一种选择
代码
#include <iostream> #include <cstdio> using namespace std; int main() { int n,k; scanf("%d%d",&n,&k); long long dp[15]; dp[1] = k-1; dp[2] = k*(k-1); for(int i = 3;i<=n;++i){ dp[i] = dp[i-1]*(k-1)+dp[i-2]*(k-1); } printf("%lld",dp ); return 0; }
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