杭电多校第十场 hdu6432 Cyclic 打表找规律

Cyclic

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 193    Accepted Submission(s): 125

Problem Description

Count the number of cyclic permutations of length n with no continuous subsequence [i, i + 1 mod n].
Output the answer modulo 998244353.

 

Input

The first line of the input contains an integer T , denoting the number of test cases.
In each test case, there is a single integer n in one line, denoting the length of cyclic permutations.
1 ≤ T ≤ 20, 1 ≤ n ≤ 100000

 

Output

For each test case, output one line contains a single integer, denoting the answer modulo 998244353.

 

Sample Input

3
4
5
6

 

Sample Output

1
8
36

 

题意：求满足一个方向的(a[i]+1)%n!=a[i+1]循环数列的n个数字组成的数列的可能性

分析：首先按照题目的意思做一个全排列列出前面几项的可能数

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = 1e5+10;
const ll mod = 998244353;
const double pi = acos(-1.0);
const double eps = 1e-8;
ll a[maxn];
int main() {
    ios::sync_with_stdio(0);
    ll T, n;
    scanf("%lld",&T);
    while( T -- ) {
        scanf("%lld",&n);
        for( ll i = 0; i < n; i ++ ) {
            a[i] = i+1;
        }
        ll cnt = 0;
        while(next_permutation(a,a+n)) {
            bool flag = true;
            for( ll i = 0; i < n-1; i ++ ) {
                ll t;
                if( (a[i]+1)%n == 0 ) {
                    t = a[i]+1;
                } else {
                    t = (a[i]+1)%n;
                }
                if( t == a[i+1] ) {
                    flag = false;
                    break;
                }
            }
            ll t;
            if( (a[n-1]+1)%n == 0 ) {
                t = a[n-1]+1;
            } else {
                t = (a[n-1]+1)%n;
            }
            if( t == a[0] ) {
                flag = false;
            }
            if(flag) {
                cnt ++;
                /*for( ll i = 0; i < n; i ++ ) {
                    cout << a[i] << " ";
                }
                cout << endl;*/
            }
        }
        printf("%lld\n",cnt/n);
    }
    return 0;
}

　　接着找规律

AC代码：

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = 1e5+10;
const ll mod = 998244353;
const double pi = acos(-1.0);
const double eps = 1e-8;
ll a[maxn];
int main() {
    ios::sync_with_stdio(0);
    ll T, n;
    a[1] = 0, a[2] = 0, a[3] = 0, a[4] = 1, a[5] = 8, a[6] = 36;
    for( ll i = 7; i <= maxn-10; i ++ ) {
        a[i] = ((i-3)*a[i-1]%mod+(i-2)*(2*a[i-2]+a[i-3])%mod)%mod;
    }
    scanf("%lld",&T);
    while( T -- ) {
        scanf("%lld",&n);
        printf("%lld\n",a[n]);
    }
    return 0;
}