Codeforces Round #131 Div1 B

Problem

给出$A_i（i\in[0,9]）$。表示每一个数字至少须要出现多少次。问有多少个数满足下面三个条件：1. 数至多有$N$位；2. 数不含有前导0；3. 每一个数 $i$ 出现的次数不少于$A_i$（mod 1e9+7）

Limits

$Time Limit(ms): 2000$

$Memory Limit(MB): 256$

$N\in[1,100]$

$A_i \in[0,100]$

Look up Original Problem From here

Solution

实际上是一种填数、计数统计的问题。

考虑填0的情况。在此基础上。考虑填1的情况。…。最后考虑填9的情况。

因为不可含有前导0，最好还是先枚举如果第一个数字，而剩下的$N－1$位数字进行填数。设$dp[i][j]$表示当前填下的这个数为 i，数的长度是 j 时有多少个数满足题意。

转移方程：$dp[now][i+j]=dp[now][i+j]+dp[now-1][j]\times C_{n-j}^i$。统计答案，$ans=\sum_{j=0}^n(\frac{dp[9][j]}{C_n^{n-j}})$。除法不满足取模运算。用逆元解决。

Complexity

$Time Complexity: O(9\cdot 9\cdot N^2)$

$Memory Complexity: O(N^2)$

My Code

//Hello. I‘m Peter.
#include<cstdio>
#include<iostream>
#include<sstream>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<cctype>
#include<ctime>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
typedef long double ld;
const double pi=acos(-1.0);
#define peter cout<<"i am peter"<<endl
#define randin srand((unsigned int)time(NULL))
#define INT (0x3f3f3f3f)*2
#define LL (0x3f3f3f3f3f3f3f3f)*2
#define input freopen("data.txt","r",stdin)
#define gsize(a) (int)a.size()
#define len(a) (int)strlen(a)
#define slen(s) (int)s.length()
#define clr(a) memset(a,0,sizeof(a))
#define clr_queue(q) while(!q.empty()) q.pop()
#define clr_stack(s) while(!s.empty()) s.pop()
#define rep(i, a, b) for (int i = a; i < b; i++)
#define dep(i, a, b) for (int i = a; i > b; i--)
#define repin(i, a, b) for (int i = a; i <= b; i++)
#define depin(i, a, b) for (int i = a; i >= b; i--)
#define esp 1e-6
#define MAXN
#define N 200
#define M
const ll mod=1e9+7;
ll C[N][N],ans,dp[15][N];
int n,a[N],sum;
void init_C(){
    clr(C);
    rep(i,0,N){
        C[i][0]=1;
    }
    rep(i,1,N){
        rep(j,1,N){
            C[i][j]=(C[i-1][j-1]+C[i-1][j])%mod;
        }
    }
}
ll quick_power(ll x,ll y){
    if(!y) return 1;
    ll res=quick_power(x,y>>1);
    res=res*res%mod;
    if(y%2) res=res*x%mod;
    return res;
}
int main(){
    init_C();
    ans=0;
    scanf("%d",&n);
    rep(i,0,10){
        scanf("%d",a+i);
    }
    repin(de,1,9){
        a[de]-=1;
        n-=1;
        repin(i,0,9){
            repin(j,0,n){
                dp[i][j]=0;
            }
        }
        repin(i,0,n){
            if(i>=a[0]) dp[0][i]=C[n][i];
        }
        repin(now,1,9){
            repin(i,0,n){
                if(i<a[now]) continue;
                repin(j,0,n){
                    if(i+j>n || !dp[now-1][j]) continue;
                    dp[now][i+j]=(dp[now][i+j]+dp[now-1][j]*C[n-j][i])%mod;
                }
            }
        }
        repin(i,0,n){
//            ans=(ans+dp[9][i]/C[n][n-i]); 不能直接求。转化为逆元来做。
            ans=(ans+dp[9][i]*quick_power(C[n][n-i],mod-2))%mod;
        }
        n+=1;
        a[de]+=1;
    }
    printf("%lld\n",ans);
}