ural 1009 记忆化搜索

1009. K-based Numbers

Time limit: 1.0 second
Memory limit: 64 MB

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

inputoutput

210

90

题意：给你一个k进制的数，让你计算出符合条件的n位的k进制的数的个数。条件就是k进制的数不能有连续的两个0，更不能有前导零。

分析：直接dfs就行，第一位要从1开始，循环到k-1，每次记录下当前的末尾的数，如果是零，那么下一位就不能取零。纯dfs是能过的400+ms，如果把计算的结果把存下来，就变成了1ms。

#include<bitset>#include<map>#include<vector>#include<cstdio>#include<iostream>#include<cstring>#include<string>#include<algorithm>#include<cmath>#include<stack>#include<queue>#include<set>#define inf 0x3f3f3f3f#define mem(a,x) memset(a,x,sizeof(a))using namespace std;typedef long long ll;typedef pair<int,int> pii;inline int in(){    int res=0;char c;    while((c=getchar())<'0' || c>'9');    while(c>='0' && c<='9')res=res*10+c-'0',c=getchar();    return res;}const int N=100100;ll dp[22][22];int n,k;ll dfs(int cnt,int tail){    if(cnt==n+1) return 1;    if(dp[cnt][tail] != -1) return dp[cnt][tail];    ll ret=0;    int mn=tail==0? 1 : 0;    for(int i=mn;i<k;i++)    {        ret += dfs(cnt+1,i);    }    return dp[cnt][tail]=ret;}int main(){    while(~scanf("%d%d",&n,&k))    {        mem(dp,-1);        printf("%I64d\n",dfs(1,0));    }    return 0;}