D. New Year and the Permutation Concatenation(打表找規律)
http://codeforces.com/contest/1091/problem/D
D. New Year and the Permutation Concatenation
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Let n
be an integer. Consider all permutations on integers 1 to n in lexicographic order, and concatenate them into one big sequence p. For example, if n=3, then p=[1,2,3,1,3,2,2,1,3,2,3,1,3,1,2,3,2,1]. The length of this sequence will be n⋅n!
.
Let 1≤i≤j≤n⋅n!
be a pair of indices. We call the sequence (pi,pi+1,…,pj−1,pj) a subarray of p. Its length is defined as the number of its elements, i.e., j−i+1. Its sum is the sum of all its elements, i.e., ∑jk=ipk
.
You are given n
. Find the number of subarrays of p of length n having sum n(n+1)2. Since this number may be large, output it modulo 998244353
(a prime number).
Input
The only line contains one integer n
(1≤n≤106
), as described in the problem statement.
Output
Output a single integer — the number of subarrays of length n
having sum n(n+1)2, modulo 998244353
.
Examples
Input
Copy
3
Output
Copy
9
Input
Copy
4
Output
Copy
56
Input
Copy
10
Output
Copy
30052700
Note
In the first sample, there are 16
subarrays of length 3
. In order of appearance, they are:
[1,2,3]
, [2,3,1], [3,1,3], [1,3,2], [3,2,2], [2,2,1], [2,1,3], [1,3,2], [3,2,3], [2,3,1], [3,1,3], [1,3,1], [3,1,2], [1,2,3], [2,3,2], [3,2,1]
.
Their sums are 6
, 6, 7, 6, 7, 5, 6, 6, 8, 6, 7, 5, 6, 6, 7, 6. As n(n+1)2=6, the answer is 9
解題思路：
比賽的時候一頭霧水。賽後打了個表也沒發現什麼規律。還是觀察力少了些。
打表可發現：
3——————9
4——————56
5——————395
6——————3084
（因為連起來的排列組合的sum一定是n*(n-1)/2,所以）
3——————9=3!+3=3!+1*3
4——————56=4!+32=4!+4*8
5——————395=5!+275=5!+5*55
6——————3084=6!+2364=6!+6*394
emmmmmmmmmm....這樣就推出來了。也是很有趣。
程式碼：
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int mod=998244353;
const int maxn=1000000+100;
LL P[maxn];
int main()
{
LL n;
while(scanf("%lld",&n)!=EOF){
LL fac;
P[1]=1;
P[2]=2;
P[3]=9;
fac=6;
for(int i=4;i<=n;i++){
fac=fac*i%mod;
P[i]=(((P[i-1]-1)%mod*i)%mod+fac)%mod;
}
printf("%lld\n",P[n]);
}
}
打表程式碼：
#include<bits/stdc++.h>
using namespace std;
int b[1000000];
int bb[1000000];
int main()
{
int n;
while(scanf("%d",&n)!=EOF){
for(int i=1;i<=n;i++){
b[i]=i;
bb[i]=i;
}
int index=n;
while(next_permutation(b+1,b+n+1)){
for(int i=1;i<=n;i++){
index++;
bb[index]=b[i];
}
}
/*cout<<"index:"<<index<<endl;
for(int i=1;i<=index;i++){
cout<<bb[i]<<" * ";
}*/
int countt=0;
for(int i=1;i<=index-n+1;i++){
int sum=0;
for(int j=i;j<i+n;j++){
sum+=bb[j];
}
///cout<<"sum:"<<sum<<endl;
if(sum==n*(n+1)/2){
countt++;
}
}
cout<<"countt:"<<countt<<endl;
cout<<endl;
}
}