【JZOJ4806】打工

description

---

analysis

• $DP$

• 易知一个数$x$若想出现，则它前面$[1,x-1]$每一个数都要出现过，否则这种情况会被顶替

• 则设$f[i][j]$表示还有$i$个空位、前面已经填过的数最大为$j$的方案数

• 就是说$f[i][j]$状态可以由填$j$个数的$j\ast f[i-1][j]$或填$j+1$的$f[i-1][j+1]$转移得到

• 最终答案就是$\sum (a[i]-1)f[n-i][max(a[1...n-1])]$

---

code

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 10005
#define mod 1000007
#define ll int
#define fo(i,a,b) for (ll i=a;i<=b;++i)
#define fd(i,a,b) for (ll i=a;i>=b;--i)

using namespace std;

ll f[MAXN][MAXN];
ll a[MAXN],mx[MAXN];
ll n,ans;

inline ll read()
{
	ll x=0,f=1;char ch=getchar();
	while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
	while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*f;
}
inline ll max(ll x,ll y){return x>y?x:y;}
int main()
{
	freopen("T1.in","r",stdin);
	n=read(),f[0][0]=1;
	fo(i,1,n)mx[i]=max(mx[i-1],a[i]=read()),f[0][i]=1;
	fo(i,1,n)
	{
		fo(j,1,n)
		f[i][j]=((long long)f[i-1][j]*j%mod+f[i-1][j+1])%mod;
	}
	fo(i,1,n)ans=(ans+(long long)(a[i]-1)*f[n-i][mx[i-1]])%mod;
	printf("%d\n",(ans+1)%mod);
	return 0;
}