[POJ 2279] Mr. Young's Picture Permutations

[题目链接]

         http://poj.org/problem?id=2279

[算法]

         杨氏矩阵与勾长公式

[代码]

       

#include <algorithm>  
#include <bitset>  
#include <cctype>  
#include <cerrno>  
#include <clocale>  
#include <cmath>  
#include <complex>  
#include <cstdio>  
#include <cstdlib>  
#include <cstring>  
#include <ctime>  
#include <deque>  
#include <exception>  
#include <fstream>  
#include <functional>  
#include <limits>  
#include <list>  
#include <map>  
#include <iomanip>  
#include <ios>  
#include <iosfwd>  
#include <iostream>  
#include <istream>  
#include <ostream>  
#include <queue>  
#include <set>  
#include <sstream>  
#include <stdexcept>  
#include <streambuf>  
#include <string>  
#include <utility>  
#include <vector>  
#include <cwchar>  
#include <cwctype>  
#include <stack>  
#include <limits.h> 
using namespace std;
#define MAXN 35
typedef long long ll;

ll i,j,k,tmp,n,cnt;
ll a[MAXN];
ll num[MAXN*MAXN];
ll x,y;

ll gcd(ll x,ll y)
{
    return y == 0 ? x : gcd(y,x % y);
}

int main()
{
    
    while (scanf("%lld",&n) != EOF && n)
    {    
        cnt = 0;
        memset(num,0,sizeof(num));
        for (i = 1; i <= n; i++) scanf("%lld",&a[i]);
        for (i = 1; i <= n; i++)
        {
            for (j = 1; j <= a[i]; j++)
            {    
                cnt++;
                for (k = i + 1; k <= n; k++)
                {
                    if (a[k] >= j) num[cnt]++;
                    else break;
                }
                num[cnt] += a[i] - j + 1;
            }
        }
        x = 1; y = 1;
        for (i = 1; i <= cnt; i++)
        {
            x *= i;
            y *= num[i];
            tmp = gcd(x,y);
            x /= tmp;
            y /= tmp;
        }
        printf("%lld
",x/y);
    }

    return 0;
}