【51nod】2589 快速讨伐

51nod 2589 快速讨伐

又是一道倒着推改变世界的题。。。

从后往前考虑，设(dp[i][j])表示还有(i)个1和(j)个(2)没有填，那么填一个1的话直接转移过来

(dp[i][j] ightarrow dp[i - 1][j])

如果填一个(2)要把(A[j])的那些敌人都扔在这个2的后面

方案是

(inom{N - i + N - j + sum[N] - sum[j - 1]}{A[j]} A[j]! dp[i][j] ightarrow dp[i][j - 1])

然后把(i < j)的状态都标成0就好了

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define eps 1e-10
#define ba 47
#define MAXN 2005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
	if(c == '-') f = -1;
	c = getchar();
    }
    while(c >= '0' && c <= '9') {
	res = res * 10 +c - '0';
	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
	out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 998244353,MAXV = 4000000;
int N;
int A[MAXN],s[MAXN];
int fac[MAXV + 5],invfac[MAXV + 5];
int dp[MAXN][MAXN];
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
void update(int &x,int y) {
    x = inc(x,y);
}
int C(int n,int m) {
    if(n < m) return 0;
    else return mul(fac[n],mul(invfac[m],invfac[n - m]));
}
int fpow(int x,int c) {
    int res = 1,t = x;
    while(c) {
	if(c & 1) res = mul(res,t);
	t = mul(t,t);
	c >>= 1;
    }
    return res;
} 
void Solve() {
    read(N);
    for(int i = 1 ; i <= N ; ++i) read(A[i]);
    s[0] = 1;
    for(int i = 1 ; i <= N ; ++i) s[i] = s[i - 1] + A[i];
    fac[0] = 1;
    for(int i = 1 ; i <= MAXV ; ++i) fac[i] = mul(fac[i - 1],i);
    invfac[MAXV] = fpow(fac[MAXV],MOD - 2);
    for(int i = MAXV - 1 ; i >= 0 ; --i) invfac[i] = mul(invfac[i + 1],i + 1);
    dp[N][N] = 1;
    for(int i = N ; i >= 0 ; --i) {
	for(int j = N ; j >= 0 ; --j) {
	    if(i < j) {dp[i][j] = 0;continue;}
	    if(i && i - 1 >= j) {
		update(dp[i - 1][j],dp[i][j]);
	    }
	    if(j) {
		update(dp[i][j - 1],mul(mul(dp[i][j],fac[A[j]]),C(N - i + N - j + s[N] - s[j - 1],A[j])));
	    }
	}
    }
    out(dp[0][0]);enter;
}
int main(){
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}