dddddd

#include<iostream>
#include<stdio.h>
using namespace std;
int record;
int N[1];


void dfs(int sum,int n)
{
   if(sum<0)
     return ;
   if(n==N[0]){
     record++;
     return;}
   dfs(sum+1,n+1);
   dfs(sum-1,n);
}   
   

int main()
{   
    record=0;
    FILE *fp;
    fp=fopen("DATA1.IN","r");
    fread(N,sizeof(int),1,fp);
    fclose(fp);
     dfs(1,1);
    N[1]=record;
     fp=fopen("ANS1.OUT","w");
    fwrite(N,sizeof(int),1,fp);
     fclose(fp);  
    return 0;
}

#include <iostream>
using namespace std;

int n;
int p = 0;
int num0 = 0,num1 = 0;

void suan(int i)
{
    if(i==2*n)
        p++;
    else
    {
         if(num0<n&&num0+1<=num1)
         {
            num0++;
            suan(i+1);
            num0--;
         }
         if(num1<n)
         {
             num1++;
             suan(i+1);
             num1--;
         }
    }
    return;
}

int main(void)
{
    FILE *f = fopen("DATA1.IN","r");
    fscanf(f,"%d",&n);
    fclose(f);
    suan(1);
    f = fopen("ANS1.OUT","w");
    fprintf(f,"%d",p);
    fclose(f);
    return 0;
}

/**********************
不懂文件输入输出 呵呵
所以只把分析出来的数学公式写出来了
具体接口麻烦thomas兄帮我修改下勒：）

以下我抽离出来的纯粹数学公式
**********/

int count(int n) //输入N  输出个数p
{
    int i = n-1;
    int j(1);
    int count(n);
    while(j<=n-2)
    {
        count=count+i*(i-1)*j;
        i--;
        j++;
    }
    return count;
}

ms是，C(N,2N)/(N+1)

受不了了。
不比速度的话，这个就是答案了。不写程序了。

#include <stdio.h>
#include <math.h>

void mulex(int n)
{
    int primeArray[]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,
                      47,53,59,61,67,71,73,79,83,89,97,101,
                      103,107,109,113,127,131,137,139,149,
                      151,157,163,167,173,179,181,191,
                      193,197,199};
    
    int primeCount1[46];
    int primeCount2[15];
    int i,j,k;
    int isPrime=0;
    int quotient;
    int res[100];
    int carry,digit=1,temp; 
    FILE *fp;
    
    primeCount1[0]=(int)((n+1)/2);
    primeCount2[0]=1;
    for(i=1;i<46;i++)
    {
        primeCount1[i]=0;
        if(i<15)primeCount2[i]=0;
    }
    primeCount2[1]=1;
    
    for(k=4;k<=n-primeCount1[0];k++)
    {    
        i=k;
        while(!isPrime)
        {
            for(j=0;j<15;j++)
            {
                if(i%primeArray[j]==0)
                {
                    quotient=i/primeArray[j];
                    primeCount2[j]++;
                    if(quotient==1)
                        isPrime=1;
                    else
                        i=quotient;
                    break;
                }
            }
        }
        isPrime=0;
    }

    for(k=primeCount1[0]*2+3;k<=2*n-1;k=k+2)
    {
        i=k;
        while(!isPrime)
        {
            for(j=0;j<46;j++)
            {
                if(i%primeArray[j]==0)
                {
                    quotient=i/primeArray[j];
                    primeCount1[j]++;
                    if(quotient==1)
                        isPrime=1;
                    else
                        i=quotient;
                    break;
                }
            }
        }
        isPrime=0;
        
    }

    for(i=0;i<15;i++)
    {
        if(primeCount2[i])
            primeCount1[i]=primeCount1[i]-primeCount2[i];
    }

    res[0]=1;

    for(i=0;i<46;i++)
    {
        if(primeCount1[i])
        {
            for(j=0;j<primeCount1[i];j++)
            {
                for(k=1,carry=0;k<=digit;k++)
                {
                    temp=res[k-1]*primeArray[i]+carry;
                    res[k-1]=temp%10;
                    carry=temp/10;
                }
                while(carry)
                {
                    res[++digit-1]=carry%10;
                    carry/=10;    
                }
            }
        }
    }

    fp=fopen("ANS1.OUT","a+");

    for(i=digit;i>=1;--i)
    {
        fprintf(fp,"%d",res[i-1]);
    }

    fclose(fp);
}

int main()
{
    
    int n;
    int i;
    unsigned long res,res1=1,res2=1;
    FILE *fp;
    
    fp=fopen("DATA1.IN","r");
    fscanf(fp,"%d",&n);
    fclose(fp);

    if(n>8)
    {
        mulex(n);
        return 0;
    }
    
    if(n)
    {
        for(i=2*n;i>n;i--)
            res1*=i;
        for(i=1;i<=n;i++)
            res2*=i;
        res=res1/res2/(n+1);
    }
    else res=0;

    fp=fopen("ANS1.OUT","a+");

    fprintf(fp,"%d",res);
    fclose(fp);
    return 0;
}

time's out[em1]

看起来象是C(n,2n)/(n+1)
但是没有证明，就不写程序了
十一大家都要出去玩的，看起来参加的人不多
thomas,不好意思催了你一下，主要是想在十一假期以前找点事做
你要是喜欢这种貌似用搜索的题目，我还能给你一道更狠的，看不出来的
呵呵