Distribute Message
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 605 Accepted Submission(s): 259
Problem Description
The contest’s message distribution is a big thing in prepare. Assuming N students stand in a row, from the row-head start transmit message, each person can transmit message to behind M personals, and how many ways could row-tail get the message?
Input
Input may contain multiple test cases. Each case contains N and M in one line. (0<=M<N<=30)
When N=0 and M=0, terminates the input and this test case is not to be processed.
When N=0 and M=0, terminates the input and this test case is not to be processed.
Output
Output the ways of the Nth student get message.
Sample Input
4 1
4 2
0 0
Sample Output
1
3
Hint
4 1 : A->B->C->D
4 2 : A->B->C->D, A->C->D, A->B->D 这题是一个简单的DP题，每一个点都将其方式保留起来，使得后面的点能够从前面得到信息。
代码如下：
1 #include <cstdio>
2 #include <cstring>
3 #include <cstdlib>
4 using namespace std;
5
6 int way[35], N, M;
7
8 /*int get( int x )
9 {
10 if( x == 1 )
11 {
12 return 1;
13 }
14 int ans = 0;
15 for( int j = 1; j <= M; ++j )
16 {
17 if( x - j >= 1 )
18 {
19 ans += get( x - j );
20 }
21 }
22 return ans;
23 } DFS超时*/
24
25 int main()
26 {
27 while( scanf( "%d %d", &N, &M ), N|M )
28 {
29 memset( way, 0, sizeof( way ) );
30 way[1] = 1;
31 for( int i = 1; i< N; ++i )
32 {
33 for( int j = 1; j <= M; ++j )
34 {
35 if( i + j > N ) break;
36 way[i+j] += way[i];
37 }
38 }
39 printf( "%d\n", way[N] );
40 }
41 return 0;
42 }