HDU - 6880 Permutation Counting

传送门：HDU - 6880

思路分析

设$dp[i][j]$表示前$i$个数且末尾为$j$的排列数
当我们将$dp[i-1][..]->dp[i][j]$的时候，实际上是看做从$i-1$的排列里把$j$换到最后一个位置，并把$i$插入到原排列且满足限制，解决办法就是将前$i-1$中大于等于$j$的数全部加一，这就就刚好空出来一个$j$，且满足限制

AC代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))

namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
    bool IOerror=0;
    inline char nc() {
        static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
        if(p1==pend) {
            p1=buf;
            pend=buf+fread(buf,1,BUF_SIZE,stdin);
            if(pend==p1) {
                IOerror=1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
    }
    template<class T> inline bool read(T &x) {
        bool sign=0;
        char ch=nc();
        x=0;
        for(; blank(ch); ch=nc());
        if(IOerror)return false;
        if(ch=='-')sign=1,ch=nc();
        for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
        if(sign)x=-x;
        return true;
    }
    template<class T,class... U>bool read(T& h,U&... t) {
        return read(h)&&read(t...);
    }
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int INF = 0x3f3f3f3f;
const int N = 5e3+10;
const int mod = 1e9+7;

#define int long long

int b[N],dp[N][N];

signed main() {

#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

    int T;
    cin>>T;
    while(T--){
        int n;
        cin>>n;
        for(int i=2;i<=n;i++) cin>>b[i];
        for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) dp[i][j]=0;
        dp[1][1]=1;
        for(int i=2;i<=n;i++){
            int sum = 0;
            //1 small
            //0 big
            if(b[i]){
                for(int j=i-1;j;j--){
                    sum = (sum + dp[i-1][j])%mod;
                    dp[i][j] = (dp[i][j]+sum)%mod;
                }
            }else{
                for(int j=2;j<=i;j++){
                    sum = (sum+dp[i-1][j-1])%mod;
                    dp[i][j] = (dp[i][j]+sum)%mod;
                }
            }
        }
        int ans = 0;
        for(int i=1;i<=n;i++) ans = (ans + dp[n][i])%mod;
        cout<<ans<<endl;
    }




    return 0;
}