CS Academy Restricted Permutations

题目链接：https://csacademy.com/contest/archive/task/restricted-permutations

题目大意：求在约束条件下1~N的全排列的个数。约束：0， 1表示的n - 1个数，1表示第i个数一定要放到第i+1个数的前面，0反之。

解题思路：我也是看的editorial。。。实际上当我们考虑第i+1个数的时候我们只需要考虑他与第i个数的关系就可以了，那么这样子我们就可以设dp[i][j]表示第i个数在位置j的时候前i个数的全排列个数。那么：

i在i-1后面：dp[i][j] = dp[i- 1][0] + dp[i - 1][1] + ... + dp[i - 1][j - 1]

i在i-1前面：dp[i][j] = dp[i - 1][i - 1] + dp[i - 1][i - 2] + ... + dp[i - 1][j + 1]

朴素写法O(n³)，可以记录dp[i- 1][0] + dp[i - 1][1] + ... + dp[i - 1][j - 1]降低到O(n²)

代码：

const int maxn = 2e3 + 5;
int sta[maxn], n;
ll dp[maxn][maxn];

void solve(){
    memset(dp, 0, sizeof(dp));
    dp[1][1] = 1; 
    ll lasum;
    for(int i = 2; i <= n; i++){
        if(sta[i]){
            lasum = dp[i - 1][0];
            for(int j = 1; j <= i; j++){
                dp[i][j] += lasum % mod;
                lasum += dp[i - 1][j] % mod;
            }
        }
        else{
            lasum = 0;
            for(int j = i; j >= 1; j--){
                dp[i][j] += lasum % mod;
                lasum += dp[i - 1][j - 1] % mod;
            }
        }
    }
    ll ans = 0;
    for(int i = 1; i <= n; i++) ans += dp[n][i] % mod;
    printf("%I64d
", ans % mod);
}

int main(){
    scanf("%d", &n);
    for(int i = 2; i <= n; i++) scanf("%d", &sta[i]);
    solv

题目：

Restricted Permutations

Time limit: 1000 ms
Memory limit: 128 MB

 

Find the number of permutations of size NN, where for each ii, 1 leq i < N1≤i<N, you know whether ii comes before or after i+1i+1.

Standard input

The first line contains a single integer NN.

The second line contains a binary array of size N-1N−1. The i^{th}i​th​​ element of this array is 11 if ii comes before i+1i+1 in the permutation, 00 otherwise.

Standard output

Print the answer modulo 10^9+710​9​​+7 on the first line.

Constraints and notes

• 2 leq N leq 20002≤N≤2000 

Input	Output	Explanation
3 1 0	2	1 2 3       1 3 2 3 1 2
5 1 1 0 1	9	1 2 3 4 5 6 7 8 9 10       1 2 4 3 5 1 2 4 5 3 1 4 2 3 5 1 4 2 5 3 1 4 5 2 3 4 1 2 3 5 4 1 2 5 3 4 1 5 2 3 4 5 1 2 3
7 1 0 1 0 1 0	272