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
Tags: dynamic programming (
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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; }
时间: 2024-10-10 13:08:24