省选模拟 19/09/13

Posted 2022-07-08 qq62c30ac77b2a7

​​组合数问题​​​ 很明显最下面的中间最大
把它放进堆，然后每次取最大，将它左右上分别放进堆中就可以了
现在的难题转换为求大小
一个套路就取 乘法转加法

#include<bits/stdc++.h>
#define N 1000050
using namespace std;
typedef long long ll;
const int Mod = 1000000007;
int read()
  int cnt = 0, f = 1; char ch = 0;
  while(!isdigit(ch)) ch = getchar(); if(ch == -) f = -1;
  while(isdigit(ch)) cnt = cnt*10 + (ch-0), ch = getchar();
  return cnt * f;

int n, k;
ll fac[N], ifac[N], a[N];
bool cmp(ll a, ll b) return a > b;
ll mul(ll a, ll b) return (a * b) % Mod;
ll add(ll a, ll b) return (a + b) % Mod;
ll power(ll a, ll b) ll ans = 1;
  for(;b;b>>=1) if(b&1) ans = mul(ans, a); a = mul(a, a);
  return ans;

ll C(int n, int m) return mul(fac[n], mul(ifac[n-m], ifac[m]));
typedef long double ld;
ld sum[N];
map<pair<int,int>, int> vis;
#define mp make_pair
struct Node 
  int x, y;
  friend bool operator < (const Node &a, const Node &b)
    return sum[a.x] + sum[b.y] + sum[b.x - b.y] < sum[b.x] + sum[a.y] + sum[a.x - a.y];
  
 b[N];
priority_queue<Node> q;
int main()
  n = read(), k = read();
  fac[0] = fac[1] = ifac[0] = ifac[1] = 1;
  for(int i = 2; i <= n; i++) fac[i] = mul(fac[i-1], i);
  ifac[n] = power(fac[n], Mod-2);
  for(int i = n-1; i >= 2; i--) ifac[i] = mul(ifac[i+1], i+1);
  sum[0] = sum[1] = log(1);
  for(int i = 2; i <= n; i++) sum[i] = sum[i-1] + log(i);
  ll ans = 0; int cnt = 0;
  q.push((Node)n, n/2);
  while(cnt < k)
    Node now = q.top(); q.pop();
    if(vis[mp(now.x, now.y)]) continue;
    vis[mp(now.x, now.y)] = 1; 
    ans = add(ans, C(now.x, now.y));
    ++cnt; 
    q.push((Node)now.x - 1, now.y);
    q.push((Node)now.x, now.y - 1);
    q.push((Node)now.x, now.y + 1);
   cout << ans; return 0;

​​字符串​​​ 首先有个暴力的莫队做法，就是先建好 ，然后每次插入删除字符串可以看做树上的链加
复杂度
注意到，于是可以把所有字符串拼在一起然后莫队，中途需要用以上是关于省选模拟 19/09/13的主要内容，如果未能解决你的问题，请参考以下文章

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