1009. K-based Numbers
Time limit: 1.0 second
Memory limit: 64 MB
Memory limit: 64 MB
Let’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
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
)
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;
}