이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#define FAST ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);cerr.tie(0)
#define mp make_pair
#define xx first
#define yy second
#define pb push_back
#define pf push_front
#define popb pop_back
#define popf pop_front
#define all(x) x.begin(),x.end()
#define ff(i,a,b) for (int i = a; i < b; i++)
#define fff(i,a,b) for (int i = a; i <= b; i++)
#define bff(i,a,b) for (int i = b-1; i >= a; i--)
#define bfff(i,a,b) for (int i = b; i >= a; i--)
using namespace std;
long double typedef ld;
unsigned int typedef ui;
long long int typedef li;
pair<int,int> typedef pii;
pair<li,li> typedef pli;
pair<ld,ld> typedef pld;
vector<vector<int>> typedef graph;
unsigned long long int typedef ull;
//const int mod = 998244353;
const int mod = 1000000007;
//Note to self: Check for overflow
int a[103];
int dp[103][103][1003][2][2];
int main()
{
FAST;
int n,L; cin>>n>>L;
fff(i,1,n) cin>>a[i];
sort(a+1,a+n+1),a[n+1]=mod;
if (n==1) return cout<<1,0;
dp[0][0][0][0][0]=1;
fff(i,1,n) fff(j,1,n) fff(s,0,L)
{
li prevsum;
prevsum=s-(li)(2*j)*(a[i+1]-a[i]);
if (prevsum>=0)
{
(dp[i][j][s][0][0]+=dp[i-1][j-1][prevsum][0][0])%=mod;
(dp[i][j][s][0][0]+=(li)2*j*dp[i-1][j][prevsum][0][0]%mod)%=mod;
(dp[i][j][s][0][0]+=(li)j*(j+1)*dp[i-1][j+1][prevsum][0][0]%mod)%=mod;
}
prevsum=s-(li)(2*j-1)*(a[i+1]-a[i]);
if (prevsum>=0)
{
(dp[i][j][s][0][1]+=dp[i-1][j-1][prevsum][0][1])%=mod;
(dp[i][j][s][0][1]+=(li)(2*j-1)*dp[i-1][j][prevsum][0][1]%mod)%=mod;
(dp[i][j][s][0][1]+=(li)(j*(j+1)-j)*dp[i-1][j+1][prevsum][0][1]%mod)%=mod;
(dp[i][j][s][0][1]+=dp[i-1][j-1][prevsum][0][0])%=mod;
(dp[i][j][s][0][1]+=(li)j*dp[i-1][j][prevsum][0][0]%mod)%=mod;
}
prevsum=s-(li)(2*j-1)*(a[i+1]-a[i]);
if (prevsum>=0)
{
(dp[i][j][s][1][0]+=dp[i-1][j-1][prevsum][1][0])%=mod;
(dp[i][j][s][1][0]+=(li)(2*j-1)*dp[i-1][j][prevsum][1][0]%mod)%=mod;
(dp[i][j][s][1][0]+=(li)(j*(j+1)-j)*dp[i-1][j+1][prevsum][1][0]%mod)%=mod;
(dp[i][j][s][1][0]+=dp[i-1][j-1][prevsum][0][0])%=mod;
(dp[i][j][s][1][0]+=(li)j*dp[i-1][j][prevsum][0][0]%mod)%=mod;
}
prevsum=s-(li)(2*j-2)*(a[i+1]-a[i]);
if (prevsum>=0)
{
(dp[i][j][s][1][1]+=dp[i-1][j-1][prevsum][1][1])%=mod;
(dp[i][j][s][1][1]+=(li)(2*j-2)*dp[i-1][j][prevsum][1][1]%mod)%=mod;
(dp[i][j][s][1][1]+=(li)(j*(j+1)-2*j)*dp[i-1][j+1][prevsum][1][1]%mod)%=mod;
(dp[i][j][s][1][1]+=dp[i-1][j-1][prevsum][0][1])%=mod;
(dp[i][j][s][1][1]+=dp[i-1][j-1][prevsum][1][0])%=mod;
(dp[i][j][s][1][1]+=(li)(j-1)*dp[i-1][j][prevsum][0][1]%mod)%=mod;
(dp[i][j][s][1][1]+=(li)(j-1)*dp[i-1][j][prevsum][1][0]%mod)%=mod;
if (i==n && j==1)
{
(dp[i][j][s][1][1]+=dp[i-1][j+1][prevsum][1][1])%=mod;
(dp[i][j][s][1][1]+=dp[i-1][j][prevsum][1][0])%=mod;
(dp[i][j][s][1][1]+=dp[i-1][j][prevsum][0][1])%=mod;
}
}
}
int ans=0;
fff(s,0,L) (ans+=dp[n][1][s][1][1])%=mod;
cout<<ans<<"\n";
}
//Note to self: Check for overflow
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