int f(int n, int m)
{
 int fn=0,i;
 if(m<0||m>10*n) return 0;
 if(n==1) return 1;
 for(i=0;i<=10;i++)
 fn+=f(n-1,m-i);
 return fn;
}

int main()
{
 int n,m;
 printf("Please enter n and m:");
 scanf("%d%d",&n,&m);
 printf("\n%d发打中%d环有%d种可能.\n",n,m,f(n,m));

 system("PAUSE");
 return 0;
} 

解法二:

 #include"stdio.h"

int main(){

int num =1,i;
for(i=1;i<=10;i++)
num = num *(9+i)/i;
printf("%d",num);
}

个人观点:
  计算机是一门和数学相关的学科,随着这门学科的发展,这种"本质"似乎被弱化了.解法1是很普通的方法,一般人都是这种思路.而解法二则是从数学角度考虑.简洁,快速.
上次去神码笔试,碰到一题
   :给定一个数,如7899,把各位数值相加7+8+9+9=33,3+3=6,用这种方法计算机下去,求最后得到的个位数

我给出的解法:
#include "stdio.h"

int main(){
   int i;
   scanf("%d",&i);
   printf("%d",i%9==0?9:i%9);
   return 0
}//~end 
 

/**
 *
 * @author ShenXiaoliang
 *
 */
public class DemoGraph {
 
 final static int VERTICE=6;
 
 private ArrayList<String> patheSet=new ArrayList<String>();
 
 private String[] ver={"a","b","c","d","e","f"};
 
 private int[][] graph=new int[6][6];
 
 private String path="";
 
 private boolean[] isVisit=new boolean[VERTICE];
 
 
 public DemoGraph() {
  
  initiGraph();
  for(int index=0;index<VERTICE;index++)
  depthSearch(index);
  show();
  
 }
 
 private void initiGraph(){
  graph[0][1]=1;
  graph[1][0]=1;
  graph[1][2]=1;
  graph[1][3]=1;
  graph[2][1]=1;
  graph[2][3]=1;
  graph[2][4]=1;
  graph[2][5]=1;
  graph[3][1]=1;
  graph[3][2]=1;
  graph[3][5]=1;
  graph[4][2]=1;
  graph[4][5]=1;
  graph[5][2]=1;
  graph[5][3]=1;
  graph[5][4]=1;
 }
 
 private void depthSearch(int start){
  
  isVisit[start]=true;
  path+=ver[start];
  if(path.charAt(path.length()-1)=='f'||path.charAt(path.length()-1)=='b') patheSet.add(path);
  for(int index=0;index<VERTICE;index++)
   if(graph[start][index]==1&&isVisit[index]==false)
    depthSearch(index);
   else continue;
  path=path.substring(0,path.length()-1);
  isVisit[start]=false;
 }
 
 
 private void show(){
  for(String pa:patheSet)
   System.out.println(pa);
 }

 /**
  * @param args
  */
 public static void main(String[] args) {
  // TODO Auto-generated method stub
  new DemoGraph();
 }

}