結果

問題 No.1258 コインゲーム
ユーザー fastmath
提出日時 2020-10-16 22:17:04
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 90 ms / 2,000 ms
コード長 2,810 bytes
コンパイル時間 1,970 ms
コンパイル使用メモリ 195,800 KB
最終ジャッジ日時 2025-01-15 08:35:09
ジャッジサーバーID
(参考情報)
judge3 / judge2
このコードへのチャレンジ
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ファイルパターン 結果
other AC * 50
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define ii pair <int, int>
#define app push_back
#define all(a) a.begin(), a.end()
#define bp __builtin_popcountll
#define ll long long
#define mp make_pair
#define f first
#define s second
#define Time (double)clock()/CLOCKS_PER_SEC
#define debug(x) std::cout << #x << ": " << x << '\n';
const int MOD = 1e9+7, N = 2007;
//need define int long long
int mod(int n) {
n %= MOD;
if (n < 0) return n + MOD;
else return n;
}
int fp(int a, int p) {
int ans = 1, c = a;
for (int i = 0; (1ll << i) <= p; ++i) {
if ((p >> i) & 1) ans = mod(ans * c);
c = mod(c * c);
}
return ans;
}
int dv(int a, int b) { return mod(a * fp(b, MOD - 2)); }
struct M {
ll x;
M (int x_) {
x = mod(x_);
}
M () {
x = 0;
}
M operator + (M y) {
int ans = x + y.x;
if (ans >= MOD)
ans -= MOD;
return M(ans);
}
M operator - (M y) {
int ans = x - y.x;
if (ans < 0)
ans += MOD;
return M(ans);
}
M operator * (M y) {
return M(x * y.x % MOD);
}
M operator / (M y) {
return M(x * fp(y.x, MOD - 2) % MOD);
}
M operator + (int y) {
return (*this) + M(y);
}
M operator - (int y) {
return (*this) - M(y);
}
M operator * (int y) {
return (*this) * M(y);
}
M operator / (int y) {
return (*this) / M(y);
}
M operator ^ (int p) {
return M(fp(x, p));
}
void operator += (M y) {
*this = *this + y;
}
void operator -= (M y) {
*this = *this - y;
}
void operator *= (M y) {
*this = *this * y;
}
void operator /= (M y) {
*this = *this / y;
}
void operator += (int y) {
*this = *this + y;
}
void operator -= (int y) {
*this = *this - y;
}
void operator *= (int y) {
*this = *this * y;
}
void operator /= (int y) {
*this = *this / y;
}
void operator ^= (int p) {
*this = *this ^ p;
}
};
M f[N], inv[N];
void prec() {
f[0] = M(1);
for (int i = 1; i < N; ++i)
f[i] = f[i - 1] * M(i);
inv[N - 1] = f[N - 1] ^ (MOD - 2);
for (int i = N - 2; i >= 0; --i)
inv[i] = inv[i + 1] * M(i + 1);
}
M C(int n, int k) {
if (n < k)
return M(0);
else
return f[n] * inv[k] * inv[n - k];
}
signed main() {
#ifdef HOME
freopen("input.txt", "r", stdin);
#else
#define endl '\n'
ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
cout.setf(ios::fixed); cout.precision(20);
#endif
prec();
int t;
cin >> t;
while (t--) {
int n, m, x;
cin >> n >> m >> x;
auto a = M(1 + m)^n;
auto b = M(1 - m)^n;
if (x == 0) {
cout << ((a + b)/2).x << endl;
}
else {
cout << ((a - b)/2).x << endl;
}
}
}
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