URAL 1009 K-based Numbers

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 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int INF = 0x3f3f3f3f;
 5 const int M = 3e5+3;
 6 
 7 int n, k;
 8 int dp[11][20][2];  // dp[i][j][k]表示i进制下j位数以k结尾的个数(k=0表示以0结尾；k=1表示不以0结尾)
 9 int main()  {
10     for( int i=2; i<=10; i++ )  {   
11         dp[i][1][0] = 0;
12         dp[i][1][1] = i-1;
13         for( int j=2; j<=18-i; j++ )    {
14             dp[i][j][0] = dp[i][j-1][1];
15             dp[i][j][1] = (dp[i][j-1][0]+dp[i][j-1][1])*(i-1);   
16         }
17     }
18     while( ~scanf("%d%d", &n, &k ) )   {
19         printf("%d\n", dp[k][n][0]+dp[k][n][1] );
20     }
21     return 0;
22 }

 

posted @ 2015-08-17 14:10  TaoTaoCome  阅读(139)  评论(0)  编辑  收藏  举报