简单的DP
4 1 4 2 0 0
1 3Hint4 1 : A->B->C->D 4 2 : A->B->C->D, A->C->D, A->B->D
#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;
}
原文:http://blog.csdn.net/zcr_7/article/details/41551063