递推DP URAL 1009 K-based Numbers

 

题目传送门

 1 /*  2  题意：n位数，k进制，求个数  3  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];  4 */  5 #include <cstdio>  6 #include <iostream>  7 #include <algorithm>  8 #include <cstring>  9 using namespace std; 10 11 const int MAXN = 22; 12 const int INF = 0x3f3f3f3f; 13 long long dp[MAXN][MAXN]; 14 15 int main(void) //URAL 1009 K-based Numbers 16 { 17 //freopen ("B.in", "r", stdin); 18 19 int n, k; 20 while (scanf ("%d%d", &n, &k) == 2) 21  { 22 memset (dp, 0, sizeof (dp)); 23 for (int i=1; i<k; ++i) dp[1][i] = 1; 24 for (int i=2; i<=n; ++i) 25  { 26 for (int j=0; j<k; ++j) 27  { 28 if (!j) 29 for (int l=1; l<k; ++l) dp[i][j] += dp[i-1][l]; 30 else 31 for (int l=0; l<k; ++l) dp[i][j] += dp[i-1][l]; 32  } 33  } 34 35 long long ans = 0; 36 for (int i=0; i<k; ++i) ans += dp[n][i]; 37 printf ("%I64d\n", ans); 38  } 39 40 return 0; 41 }

 

 1 /*

 2  题意：n位数，k进制，求个数  3  递推DP：dp[i] = (dp[i-1] + dp[i-2]) * (k-1);  4 */

 5 #include <cstdio>

 6 #include <iostream>

 7 #include <algorithm>

 8 #include <cstring>

 9 using namespace std; 10 

11 const int MAXN = 22; 12 const int INF = 0x3f3f3f3f; 13 long long dp[MAXN]; 14 

15 int main(void)        //URAL 1009 K-based Numbers

16 { 17     //freopen ("B.in", "r", stdin);

18 

19     int n, k; 20     while (scanf ("%d%d", &n, &k) == 2) 21  { 22         memset (dp, 0, sizeof (dp)); 23         

24         dp[0] = 1;    dp[1] = k - 1; 25         for (int i=2; i<=n; ++i) 26  { 27             dp[i] = (dp[i-1] + dp[i-2]) * (k-1); 28  } 29 

30         printf ("%I64d\n", dp[n]); 31  } 32 

33     return 0; 34 }

另一种思路