代码

// This code is AI-generated. (AI 生成的代码)
// NOIP2018 填数游戏 (enhanced): closed-form answer by min(n,m) and the gap.
#include <stdio.h>
typedef long long ll;
static const ll MOD = 1000000007;
static ll pw(ll a, ll b) { ll r = 1 % MOD; a %= MOD; while (b) { if (b & 1) r = r * a % MOD; a = a * a % MOD; b >>= 1; } return r; }
int main() {
    ll n, m;
    scanf("%lld%lld", &n, &m);
    if (n > m) { ll t = n; n = m; m = t; }
    if (n == 1) { printf("%lld\n", pw(2, m)); return 0; }
    if (n == 2) { printf("%lld\n", 4 * pw(3, m - 1) % MOD); return 0; }
    if (n == 3) { printf("%lld\n", 112 * pw(3, m - 3) % MOD); return 0; }
    ll p2n = pw(2, n);
    ll ans1 = pw(4, n - 2) * p2n % MOD;
    ll ans2 = 5 * pw(4, n - 4) % MOD * p2n % MOD;
    ll inv3 = pw(3, MOD - 2);
    ll ans3 = 20 * p2n % MOD * ((pw(4, n - 4) - 1 + MOD) % MOD) % MOD * inv3 % MOD;
    ans3 = (ans3 + 15 * pw(2, n - 1)) % MOD;
    ll ans = ((ans1 + ans2) % MOD + ans3) % MOD * 2 % MOD;
    if (n == m) { printf("%lld\n", ans); return 0; }
    ans = 3 * ((ans - p2n) % MOD + MOD) % MOD;
    ans = ans * pw(3, m - n - 1) % MOD;
    printf("%lld\n", ans);
    return 0;
}

评测结果