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>
using namespace std;
typedef long long ll;
#define FOR(i, a, b) for(int i = (a); i < (b); i++)
#define F0R(i, x) FOR(i, 0, x)
const int MAX_N = 102;
const int MAX_L = 1002;
const ll MOD = 1000000007;
ll ad(ll& a, ll b, ll mod = MOD){
a = a+b; if(a >= mod) a -= mod;
return a;
}
int n, l;
ll dp[MAX_N][MAX_N][MAX_L][3], a[MAX_N];
int main(int argc, const char * argv[]) {
cin >> n >> l; F0R(i, n) cin >> a[i];
sort(a, a+n), a[n] = 10001;
if(n == 1){ cout << 1 << endl; return 0; }
if(a[1]-a[0] <= l) dp[1][1][a[1]-a[0]][1] = 2;
if(2*(a[1]-a[0]) <= l) dp[1][1][2*(a[1]-a[0])][0] = 1;
FOR(i, 1, n) FOR(j, 1, i+1) F0R(k, l+1) F0R(z, 3) if(dp[i][j][k][z]){
int d = a[i+1]-a[i];
if(z < 2 && k+d*(2*j-z-1) <= l) ad(dp[i+1][j][k+d*(2*j-z-1)][z+1], (2-z)*dp[i][j][k][z]);
if(z < 2 && k+d*(2*j-z+1) <= l) ad(dp[i+1][j+1][k+d*(2*j-z+1)][z+1], (2-z)*dp[i][j][k][z]);
if(k+d*(2*j-z) <= l) ad(dp[i+1][j][k+d*(2*j-z)][z], (2*j-z)*dp[i][j][k][z]);
if(k+d*(2*j-z+2) <= l) ad(dp[i+1][j+1][k+d*(2*j-z+2)][z], (j-z+1)*dp[i][j][k][z]);
if(j && k+d*(2*j-z-2) <= l) ad(dp[i+1][j-1][k+d*(2*j-z-2)][z], (j-1)*dp[i][j][k][z]);
}
ll ans = 0; F0R(i, l+1) ad(ans, dp[n][1][i][2]);
cout << ans << endl;
return 0;
}
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