递推DP URAL 1009 K-based Numbers

 

题目传送门

题意：n位数，k进制，求个数
分析：dp[i][j] 表示i位数，当前数字为j的个数；若j==0，不加dp[i-1][0];

 

代码1：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;

const int MAXN = 22;
const int INF = 0x3f3f3f3f;
long long dp[MAXN][MAXN];

int main(void)        //URAL 1009 K-based Numbers
{
    //freopen ("B.in", "r", stdin);

    int n, k;
    while (scanf ("%d%d", &n, &k) == 2)
    {
        memset (dp, 0, sizeof (dp));
        for (int i=1; i<k; ++i)    dp[1][i] = 1;
        for (int i=2; i<=n; ++i)
        {
            for (int j=0; j<k; ++j)
            {
                if (!j)
                    for (int l=1; l<k; ++l)    dp[i][j] += dp[i-1][l];
                else
                    for (int l=0; l<k; ++l)    dp[i][j] += dp[i-1][l];
            }
        }

        long long ans = 0;
        for (int i=0; i<k; ++i)    ans += dp[n][i];
        printf ("%I64d\n", ans);
    }

    return 0;
}

 

代码2(空间优化)：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;

const int MAXN = 22;
const int INF = 0x3f3f3f3f;
long long dp[MAXN];

int main(void)        //URAL 1009 K-based Numbers
{
    //freopen ("B.in", "r", stdin);

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

        printf ("%I64d\n", dp[n]);
    }

    return 0;
}