[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 ==  ? x : gcd(y,x % y);
}

int main()
{

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

    return ;
}