简单的DP

Distribute Message

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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.

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

Author

威士忌

#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
#include <limits.h>
#include <ctype.h>
#include <string.h>
#include <string>
#include <math.h>
#include <queue>
#include <stack>
#include <algorithm>
#include <iostream>
#include <deque>
#include <vector>
#include <set>
#include <map>
using namespace std;
#define MAXN 35
int dp[MAXN];

int main(){
    int n,m;
    int i,j;

    while(~scanf("%d%d",&n,&m)){
        if(n==0 && m==0){
            break;
        }
        memset(dp,0,sizeof(dp));
        dp[1] = 1;
        for(i=1;i<=n;i++){
            for(j=1;j<=m;j++){
                if((i+j)<=n){
                    dp[i+j] = dp[i+j]+dp[i];
                }
            }
        }
        printf("%d\n",dp[n]);
    }

    return 0;
}