bzoj 1005 HNOI2008 明明的烦恼

    这题做的我欲哭无泪啊……

    我已经没心情多说啥了……

    高精度T啊！我太弱啊！改了一天啊！还不如滚粗啊！

    想好式子在写啊！

    能用高精度乘单精度就不要用高精度乘高精度啊！

    能用高精度除单精度就不要用高精度除高精度啊！

    T成翔啊！

    简单的组合计数问题啊！

    题解在这里=_= : http://www.cnblogs.com/zhj5chengfeng/archive/2013/08/23/3278557.html

    上代码：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cmath>
#define N 1100
using namespace std;
 
class bignum
{
    public:
        int l;
        int num[N*3];
        bignum(int nownum)
        {
            memset(num, 0, sizeof(num)); l = 0;
            if (nownum == 0) l = 1;
            else while (nownum)
            {
                num[++l] = nownum % 10;
                nownum /= 10;
            }
        }
        void jia(int y);
        void chu(int y);
        void cheng(int y);
        void print();
};
int n, d[N], sum = 0;
 
void bignum::jia (int y)
{
    num[1] += y; int d = num[1] / 10; int i = 1;
    while (d)
    {
        num[++i] = d;
        d = num[i] / 10;
        num[i] %= 10;
    }
    if (i > l) l = i;
}
 
void bignum::cheng (int y)
{
    for (int i = 1; i <= l; ++i)
        num[i] = num[i] * y;
    int d = 0;
    for (int i = 1; i <= l; ++i)
    {
        num[i] += d;
        d = num[i] / 10;
        num[i] %= 10;
    }
    while (d)
    {
        num[++l] = d;
        d = num[l] / 10;
        num[l] %= 10;
    }
    while (l > 1 && num[l] == 0) l--;
}
 
void bignum::chu (int y)
{
    int yu = 0;
    for (int i = l; i >= 1; --i)
    {
        yu = yu*10 + num[i];
        num[i] = yu / y;
        yu %= y;
    }
    while (l > 1 && num[l] == 0) l --;
}
 
void bignum::print()
{
    int d = 0;
    for (int i = 1; i <= l; ++i)
    {
        num[i] += d;
        d = num[i] / 10;
        num[i] %= 10;
    }
    while (d) { num[++l] += d; d = num[l] / 10; num[l] %= 10; }
    for (int i = l; i >= 1; --i) printf("%d", num[i]);
    printf("\n"); return;
}
 
int main()
{
    scanf("%d", &n); int fre = 0;
    for (int i = 1; i <= n; ++i)
    {
        scanf("%d", &d[i]);
        if (d[i] != -1) { fre++; sum += d[i]-1; }
    }
    if (sum > n-2)
    {
        printf("0\n");
        return 0;
    }
    bignum ans = 1;
    for (int i = n-1-sum; i <= n-2; ++i) ans.cheng(i);
    for (int i = 1; i <= n-sum-2; ++i) ans.cheng(n-fre);
    for (int i = 1; i <= n; ++i)
        if (d[i] > 2)
            for (int j = 2; j < d[i]; ++j) ans.chu(j);
    ans.print();
    return 0;
}