51Nod-1799-二分答案
阿新 • • 發佈:2018-12-23
描述
題解
我們知道二分的過程中,不管是否有序,對二分結果產生直接影響的是 mid 位置的資料,而其他無關位置的資料就隨意一些了,所以我們只需要先通過一邊二分求出左右 l 和 r 的更新次數,然後求相關的排列,最後這些無關的位置只是一個階乘罷了。這裡問題也就出在了階乘,並沒有什麼特別高效的處理階乘的手法,所以這裡鑑於 n 不是特別大,所以我們可以先寫一個程式打一下表,預處理一下……
當然,不是要我們把
程式碼
#include <iostream>
using namespace std;
typedef long long ll;
const int MAG = 1e7;
const int MOD = 1e9 + 7;
const int TMP[] = {1, 682498929, 491101308, 76479948, 723816384, 67347853, 27368307, 625544428, 199888908, 888050723, 927880474, 281863274, 661224977, 623534362, 970055531, 261384175 , 195888993, 66404266, 547665832, 109838563, 933245637, 724691727, 368925948, 268838846, 136026497, 112390913, 135498044, 217544623, 419363534, 500780548, 668123525, 128487469, 30977140, 522049725, 309058615, 386027524, 189239124, 148528617, 940567523, 917084264, 429277690, 996164327, 358655417, 568392357, 780072518, 462639908 , 275105629, 909210595, 99199382, 703397904, 733333339, 97830135, 608823837, 256141983, 141827977, 696628828, 637939935, 811575797, 848924691, 131772368, 724464507, 272814771, 326159309, 456152084, 903466878, 92255682, 769795511, 373745190, 606241871, 825871994, 957939114, 435887178, 852304035, 663307737, 375297772, 217598709, 624148346, 671734977, 624500515, 748510389, 203191898, 423951674, 629786193, 672850561, 814362881, 823845496, 116667533, 256473217, 627655552, 245795606, 586445753, 172114298, 193781724, 778983779, 83868974, 315103615, 965785236, 492741665, 377329025, 847549272, 698611116};
ll fac(ll x)
{
if (x == 0)
{
return 1;
}
ll res = TMP[x / MAG];
for (ll i = x / MAG * MAG + 1; i <= x; i++)
{
res = (res * i) % MOD;
}
return res;
}
int main(int argc, const char * argv[])
{
ll n, m, k;
while (cin >> n >> m >> k)
{
ll l = 1, r = n, mid = (l + r) / 2, lcnt = 0, rcnt = 0;
while (l <= r)
{
if (mid <= k)
{
l = mid + 1;
lcnt++;
}
else
{
r = mid - 1;
rcnt++;
}
mid = (l + r) / 2;
}
ll res = 1;
for (ll i = m - lcnt + 1; i <= m; i++)
{
res *= i;
res %= MOD;
}
for (ll i = n - m - rcnt + 1; i <= n - m; i++)
{
res *= i;
res %= MOD;
}
res = res * fac(n - lcnt - rcnt) % MOD;
cout << res << '\n';
}
return 0;
}