[HAOI 2010] 计数

[题目链接]

https://www.lydsy.com/JudgeOnline/problem.php?id=2425

[算法]

类似与数位动态规划的思想 ， 用组合数学进行简单推导即可

时间复杂度 : O(L ^ 3)

[代码]

#include<bits/stdc++.h>
using namespace std;
#define N 110
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

#define int ll

int L , L0;
int tmp[N] , digit[N] , cnt[N];
char s[N];

template <typename T> inline void chkmin(T &x , T y) { x = min(x , y); }
template <typename T> inline void chkmax(T &x , T y) { x = max(x , y); }
template <typename T> inline void read(T &x)
{
   T f = ; x = ;
   char c = getchar();
   for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
   for (; isdigit(c); c = getchar()) x = (x << ) + (x << ) + c - '';
   x *= f;
}
inline ll quickpow(ll a , int n)
{
    ll b = a , res = ;
    while (n > )
    {
        if (n & ) res *= b;
        b = b * b;
        n >>= ;
    }
    return res;
}
inline void add(int x , int val)
{
    for (int i = ; i <= (int)sqrt(x); i++)
    {
        if (x % i == )
        {
            while (x % i == )
            {
                tmp[i] += val;
                x /= i;
            }
        }
    }
    if (x > ) tmp[x] += val;
}

signed main()
{

    scanf("%s" , s + );
    L = strlen(s + );
    for (int i = ; i <= L; i++)
    {
        if (s[i] > '') ++L0;
        ++cnt[s[i] - ''];
        digit[i] = s[i] - '';
    }
    ll ans = ;
    for (int k = ; k < L;k++)
    {
        if (k < L0) continue;
        memset(tmp ,  , sizeof(tmp));
        for (int i = ; i <= L0; i++) add(i , );
        for (int i = ; i <= ; i++)
        {
            for (int j = ; j <= cnt[i]; j++)
            {
                add(j , -);
            }
        }
        for (int i = ; i <= k - ; i++) add(i , );
        for (int i = ; i <= L0 - ; i++) add(i , -);
        for (int i = ; i <= k - L0; i++) add(i , -);
        ll cont = ;
        for (int i = ; i <= ; i++) cont *= quickpow(i , tmp[i]);
        ans += cont;
    }
    for (int i = ; i <= L; i++)
    {
        for (int j = ; j < digit[i]; j++)
        {
            if (i ==  && !j) continue;
            if (!j || cnt[j] > )
            {
                --cnt[j];
                int nowcnt = ;
                for (int k = ; k <= ; k++) nowcnt += cnt[k];
                if (nowcnt > L - i)
                {
                    ++cnt[j];
                    continue;
                }
                memset(tmp ,  , sizeof(tmp));
                for (int k = ; k <= nowcnt; k++) add(k , );
                for (int x = ; x <= ; x++)
                {
                    for (int y = ; y <= cnt[x]; y++)
                    {
                        add(y , -);
                    }
                }
                for (int k = ; k <= L - i; k++) add(k , );
                for (int k = ; k <= nowcnt; k++) add(k , -);
                for (int k = ; k <= L - i - nowcnt; k++) add(k , -);
                ll cont = ;
                for (int k = ; k <= ; k++) cont *= quickpow(k , tmp[k]);
                ans += cont;
                ++cnt[j];
            }
        }
        --cnt[digit[i]];
    }
    printf("%lld\n" , ans);

    return ;
}