Submission #19975

#	Time	Username	Problem	Language	Result	Execution time	Memory
19975		tonyjjw	로봇 (kriii4_F)	C++14	4 / 100	95 ms	2824 KiB

F

#include <algorithm>
#include <assert.h>
#include <complex>
#include <ctype.h>
#include <functional>
#include <iostream>
#include <limits.h>
#include <locale.h>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <vector>
#include <unordered_set>
#include <unordered_map>

#pragma warning(disable:4996)
using namespace std;

#define mp make_pair
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair <ll, int> pli;
typedef pair <ldb, ldb> pdd;

int IT_MAX = 131072;
const ll MOD = 1000000007;
const int INF = 1034567891;
const ll LL_INF = 2234567890123456789ll;
const db PI = 3.141592653589793238;
const ldb ERR = 1E-10;

ll mul_inv(ll a, ll b)
{
	ll b0 = b, t, q;
	ll x0 = 0, x1 = 1;
	if (b == 1) return 1;
	while (a > 1) {
		q = a / b;
		t = b, b = a % b, a = t;
		t = x0, x0 = x1 - q * x0, x1 = t;
	}
	if (x1 < 0) x1 += b0;
	return x1;
}

ll dp[21][41][41][4];
int P[4][2] = { { 1, 0 }, { 0, 1 }, { -1, 0 }, { 0, -1 } };

int main() {
	int N, L, M, R, i, j, k, l;
	scanf("%d %d %d %d", &N, &L, &M, &R);

	dp[0][20][20][0] = 1;
	for (i = 0; i < N; i++) {
		for (j = 1; j <= 39; j++) {
			for (k = 1; k <= 39; k++) {
				for (l = 0; l < 4; l++) {
					int d = l;
					int x = j + P[d][0];
					int y = k + P[d][1];
					dp[i + 1][x][y][d] += dp[i][j][k][l] * M % MOD * mul_inv(L + M + R, MOD);
					dp[i + 1][x][y][d] %= MOD;

					d = (l + 1) % 4;
					x = j + P[d][0];
					y = k + P[d][1];
					dp[i + 1][x][y][d] += dp[i][j][k][l] * L % MOD * mul_inv(L + M + R, MOD);
					dp[i + 1][x][y][d] %= MOD;

					d = (l + 3) % 4;
					x = j + P[d][0];
					y = k + P[d][1];
					dp[i + 1][x][y][d] += dp[i][j][k][l] * R % MOD * mul_inv(L + M + R, MOD);
					dp[i + 1][x][y][d] %= MOD;

				}
			}
		}
	}

	ll ans = 0;
	for (i = 0; i <= 40; i++) for (j = 0; j <= 40; j++) for (k = 0; k < 4; k++) ans = (ans + ((i - 20)*(i - 20) + (j - 20)*(j - 20))*dp[N][i][j][k]) % MOD;
	printf("%lld\n", ans);
	return 0;
}

• Subtask 1: 4 / 4
• Subtask 2: 0 / 96