zoukankan      html  css  js  c++  java
  • 「PKUSC2018」真实排名

    分类讨论这个数((a_i))翻不翻倍。

    若翻倍,则( ext{[)a_i$ ,2(a_i) -1]}(内的数必须翻倍,要不然会影响答案。 若不翻倍,则) ext{[(lceil frac{a_i}{2} ceil), (a_i-1)]}$内的数必须不翻倍,要不然也会影响答案。

    剩下的计算可以组合数解决,注意特判(a_i=0)的时候要输出(C_n^k)

    #include <bits/stdc++.h>
    
    using namespace std;
    typedef long long LL;
    
    #define LOG(...) fprintf (stderr, __VA_ARGS__)
    #define pb push_back
    #define mp make_pair
    #define SZ(x) ((int)(x).size())
    #define ALL(x) (x).begin(), (x).end()
    
    const int INF = 0x3f3f3f3f, MOD = 998244353;
    const LL INFL = 0x3f3f3f3f3f3f3f3fll;
    
    const int N = 100005; 
    int n, k, a[N], b[N], fac[N], ifac[N]; 
    int binom (int n, int m) {
    	if (n < m || n < 0 || m < 0) return 0; 
    	return 1LL * fac[n] * ifac[n - m] % MOD * ifac[m] % MOD;
    }
    int inv (int x) {
    	int t = MOD - 2, res = 1; 
    	while (t) {
    		if (t&1) res = 1LL * res * x % MOD;
    		x = 1LL * x * x % MOD;
    		t >>= 1; 
    	}
    	return res; 
    }
    void work() {
    	fac[0] = 1; 
    	for (int i = 1; i <= n; i++) fac[i] = 1LL * fac[i - 1] * i % MOD;
    	ifac[n] = inv(fac[n]);
    	for (int i = n-1; i >= 0; i--) ifac[i] = 1LL * ifac[i + 1] * (i + 1) % MOD;
    }
    int Q (LL x) {
    	if (x < b[1]) return 0;
    	int l = 1, r = n, ans = 0;
    	while (l <= r) {
    		int mid = (l + r) >> 1;
    		if (b[mid] <= x) ans = mid, l = mid + 1;
    		else r = mid - 1;
    	}
    	return ans; 
    }
    
    int add (int a, int b) {
    	int val = a + b;
    	if (val >= MOD) val -= MOD;
    	return val; 
    }
    int main() {
    #ifdef LiM_817
    	freopen ("input.txt", "r", stdin);
    //	freopen ("test.out", "w", stdout);
    	double cl = clock();
    #endif
    
    	cin >> n >> k;
    	for (int i = 1; i <= n; i++) scanf ("%d", &a[i]), b[i] = a[i];
    	sort (b + 1, b + n + 1);
    	work();
    	
    	for (int i = 1; i <= n; i++) {
    		if (!a[i]) {
    			printf ("%d
    ", binom (n, k)); 
    			continue;
    		}
    		
    		int ans = 0; 
    		// case 1
    		int Lb = a[i] / 2 + (a[i] % 2), Rb = a[i] - 1;
    		if (Lb > Rb) ans = add(ans, binom (n - 1, k));
    		else {
    			int q = Q(Rb) - Q(Lb - 1);
    			ans = add(ans, binom (n - 1 - q, k));
    		}
    		
    		Lb = a[i], Rb = 2 * a[i] - 1;
    		if (Lb > Rb) ans = add (ans, binom (n - 1, k - 1));
    		else {
    			int q = Q(Rb) - Q(Lb - 1);
    			ans = add (ans, binom (n - q, k - q));
    		}
    		printf ("%d
    ", ans);
    	}
    
    #ifdef LiM_817
    	cerr << fixed << setprecision(2) << "Time ellapsed: " << ((double)(clock() - cl) / CLOCKS_PER_SEC) << "s
    ";
    #endif
    	return 0;
    }
    
  • 相关阅读:
    完整的CRUD——javaweb
    JDBC预编译语句表名占位异常
    自定义标签2.x
    实现一个自定义<table>标签
    带动态属性的自定义标签
    自定义标签库_tag
    poj 3783
    The North American Invitational Programming Contest 2018 E. Prefix Free Code
    hdu 6383
    ACM-ICPC 2016 Qingdao Preliminary Contest G. Sort
  • 原文地址:https://www.cnblogs.com/LiM-817/p/11903483.html
Copyright © 2011-2022 走看看