Binary Search

链接 : http://codeforces.com/problemset/problem/1436/C

标签 二分 *1500

写了之后觉得是一道很好的用来理解二分过程的一道题.

思路 :

实际上就是模拟二分的过程, 想在乱序的序列中找到某个x, 实际上只要控制每次二分的mid值.

分别用a, b来表示用来控制二分过程的比x小的值与比x大的值.

代码 :

#include <bits/stdc++.h>
using namespace std;
#define IO ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
inline int lowbit(int x) { return x & (-x); }
#define ll long long
#define pb push_back
#define PII pair<int, int>
#define fi first
#define se second
#define inf 0x3f3f3f3f
const int N = 1010, p = 1e9 + 7;
int C[N][N];

void init() {
   for (int i = 0; i < N; ++i) 
      for (int j = 0; j <= i; ++j) {
         if (!j) C[i][j] = 1;
         else C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % p;
     }
}

int main() {	
    IO;
    init();
    int n, x, pos, a = 0, b = 0;
    cin >> n >> x >> pos;
    int l = 0, r = n;
    while (l < r) {
        int mid = l + r >> 1;
        //cout << mid << endl;
        if (mid > pos) {
            b++;
            r = mid;
        } else {
            if (mid != pos) a++;
            l = mid + 1;
        }
    }
    //cout << x - 1 << " " << a << " " << n - x << " " << b << endl;
    if (x - 1 < a || n - x < b) {
        cout << "0" << endl;
        return 0;
    }
    ll ans = 1;
    for (int i = x - 1; i > x - 1 - a; --i) ans = ans * i % p;
    for (int i = n - x; i > n - x - b; --i) ans = ans * i % p;
    if (x == 1 || x == n) {
        a = b = 0;
        ans = 1;
    }
    for (int i = 1; i <= n - a - b - 1; ++i)  ans = ans * i % p;
    cout << ans << endl;
    return 0;
}

Binary Search

原文：https://www.cnblogs.com/phr2000/p/13942697.html